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Physics 30 Electric Forces and Fields

S Molesky @ Notre Dame Page 1

Worksheet 1: Electrostatics

1) Explain why it is electrons and not protons which are

thought to be exchanged in electrostatic interactions.

Electrons are located on the exterior of the atom, while the positive

protons are located in the nucleus. In frictional charging

only the exterior parts of the atoms will be in contact. The

protons are also protected by layers of electrons.

2) How many electrons make up a charge of 100 µC?

(6.25 x 1014 e-)

4 14

19

1 e1.00 x 10 C x = 6.25 x 10 e

1.60 x 10 C

−− −

−

3) The number of electrons (elementary charges) in a coulomb

is 6.25 x 1018. If you could count 4.0 electrons per second,

and if you worked 8.0 h each of the 365 days of a year, how

many years would it take to count this number of electrons ?

(1.5 x 1011 a)

18

3

11

1 s 1 h6.25 x 10 e x x

4 e 3.6 x 10 s

1 d 1 ax x = 1.5 x 10 a

8 h 365 d

−

−

4) What is the charge on a metal sphere if it has:

a) an excess of 1.0 x 1015 electrons?

b) a deficiency of 1.0 x 1015 electrons?

( -1.6 x 10-4 C)

1915 41.60 x 10 C

1.0 x 10 e x = -1.6 x 10 C1 e

−− −

−

1915 41.60 x 10 C

1.0 x 10 e x = +1.6 x 10 C1 e

−− −

−

5) A strip of acetate and a strip of silk are rubbed together.

What can be said about the charges before and after the |

interaction?

The same number of charges are present after the interaction

as before. Charge is conserved. The negative electrons are

merely separated from one atom and attracted to another.

6) What property makes a metal a good conductor and rubber

a good insulator?

Metals have free mobile electrons whereas insulators have

localized electrons.

7) A metal sphere and a plastic sphere of the same size are

supported on insulated stands. Each is now touched to a

charged body. Compare the distribution of charge on each

sphere.

In the metal, the charge will reside evenly on the exterior

surface as electrons repel one another. In the insulator, the

charge will be localized where the object was touched, but

will also reside on the exterior surface.

8) Why do certain types of clothing tend to stick together

when they are removed from a clothes dryer?

In the dryer they are charge by frictional contact. Since they

are insulators, they do not ground out to the metal surface of

the dryer.

9) What is the significance of Benjamin Franklin (1706 - 1790)

in the study of electrostatics? Why was Franklin's designation

of charges as positive and negative better than the terms vitreous

and resinous? What problem did Franklin inadvertently create?

Franklin coined the terms positive and negative and invented a

one fluid theory that more clearly explained what happens in

static charging. He performed numerous electrostatic

experiments including the famous kite-flying experiment that

showed that lightning is an electrostatic phenomena. In designating

excess charge as positive, Franklin started the convention that

positive charges move in an electric field, which gives the

impression that it is protons, not electrons, which are electrically|

transferred.



10) Explain why trucks carrying flammable fluids drag a

chain along the ground?

The chain acts as a ground to dissipate any static electric

charge which might build up on the truck as a result of

frictional charging (metal to air or rubber to ground)

11) Why is there a limit to the quantity of charge which can

be acquired by a body?

As an object becomes more negative (or positive) it will start

to repel like charges and attract opposite charges. This will

change the objects ability to attract further similar charge. A

very large charge will also spontaneously discharge through

the air to the ground.

12) If you charge a pocket comb by rubbing with a silk scarf,

how can you determine if the comb is positively or negatively

charged?

You can test the charge on any object by bringing it close to

an electroscope of known charge. If the leaves start to repel

further, the object has the same charge as the electroscope, and

if the leaves start to drops, it has the opposite charge as the

electroscope.

13) If a strongly negatively charged body is brought near the

terminal of a charged electroscope, the leaves come together.

When the rod is brought still closer, but not touching, the

leaves again diverge. Explain why this might happen.

The induced charge separation will push negative charges

down into the leaves of a positively charged electroscope. If

the object has a sufficiently large charge, it could push so

many electrons down that the leaves actually become

negatively charged and then start to repel one another again.

14) Why does a plastic ruler that has been rubbed with a cloth

have the ability to pick up small pieces of paper? Why is it

difficult to do on a humid day?

The plastic ruler (negatively charged) will cause an induced

charge separation in the molecules of the insulator paper. The

upper surface of the paper will thus be positively charged and

since unlike charges attract, it will be pulled toward the plastic

ruler. (the unlike charges being at a shorter distance will have

a stronger force of attraction than the like negative charges which

repel at a greater distance.) One a humid day, water molecules

in the air will be attracted to charged bodies and cause an

increased gravitational force which works against the electrical

attraction.

15) Can you suggest an experiment that will prove the law

of conservation of charge?

Faraday’s Ice Bucket Experiment shows that charge is

conserved in frictional interactions

16) In charging an electroscope by induction, why is the

finger withdrawn first, then the rod?

If the rod is removed before the finger, charges will again

flow through the ground to neutralize the electroscope and

as a result, there will be no charge on the electroscope.

17) Two identical metal spheres have charges of q1 and q2.

They are brought together so they touch, and then they are

separated.

a) How is the net charge on the two spheres before they

touch related to the net charge after they touch?

The net charge before and after charging will be the same.

Charges are merely rearranged, they are neither created nor

destroyed.

b) After they touch and are separated, is the charge on each

sphere the same? Why?

Since they are conductors, the charge will be equal on the two

spheres and will be located on the exterior of the spheres.

18) Four identical metal spheres have charges of

qA = -8.0 µC, qB = -2.0 µC, qC = +5.0 µC, and

qD = +12.0 µC.

a) Two of the spheres are brought together so they touch

and then they are separated. Which spheres are they, if the

final charge on each one is +5.0 µC?

charge 1 + charge 2Since = +5 C

2

charge 1 + charge 2 = 10 C which

would indicate charges B and D

µ

µ

b) In a similar manner, which three spheres are brought

together and then separated, if the final charge on each

one is +3.0 µC? c) How many electrons would have to be

added to one of the spheres in part b) to make it electrically

neutral?

charge 1 + charge 2 + charge 3Since = +3 C

3

charge 1 + charge 2 + charge 3 = 9 C which

would indicate charges A, C and D

µ

µ



19) Describe what happens at each of the stages in charging

a neutral electroscope negatively using a positively charged

glass rod.

If an opposite charge is permanently given to the electroscope,

charge by induction is indicated. In the first stage a negative

charge is brought near the electroscope and an induced charge

separation occurs on the electroscope in which the leaves and

far side of the ball are negatively charged and the near side of the

ball is positively charged. Then the ball is momentarily grounded

on the far side and electrons leave to the ground. When the

ground is removed, electrons cannot return. When the charging

body is subsequently removed, there is a net positive charge on

the electroscope. The leaves and all surfaces of the electroscope

are positively charged.

20) Explain why the force of repulsion is the only sure test

of the sign of an electric charge.

Two bodies could be attracted either by an induced charge

separation or by unlike charges, but can only be repelled because

the charges are the same. If two bodies attract, you can’t tell

whether the second body was neutral or of an unlike charge.

21) A charged body when brought near a suspended pith ball

will first attract it. If the charged body and the pith ball then

touch, the ball will be repelled. Explain.

In the first case, there is an induced charge separation on the pith

ball whereby the far side of the ball has a like charge and the near

side has an unlike charge due to the motion of electrons. As it is a

conductor, electrons are free to move throughout the surface of the

pith ball. The near side is at a shorter distance, therefore the

attractive forces of the unlike charges will be stronger than the

repulsive forces of the like charges. Upon touching (conduction),

electrons will be transferred from the more negative body to the

less negative body and the like charges will subsequently repel.

Worksheet 2: Applications of Charging

1) Completely explain the process by which an electrostatic

precipitator removes fly ash from industrial smoke. Include

an appropriate diagram.

In the electrostatic precipitator, the large ash particles are

first removed by passing through a mechanical filter such as

asbestos. The smaller particle then pass through an

electrically charged grid which charges the small ash

particles with a coronal discharge from the closely placed

metallic mesh. The particles are then removed from the air

when they pass through a second grid with opposite charge.

As unlike charges attract, the ash will stick to the second grid,

and can be removed by washing the grid at some later time.

2) State the 5 steps involved in producing a Xerox copy.

Specify the types of charging involved at each step (where

appropriate)

1) The selenium drum is charged one its surface through a

coronal discharge from the corotron. (small sharp points

which pass charge to the drum.

2) A halogen light reflects from the ‘to be copied’ surface to

the drum and discharges the drum wherever the light contacts

(i.e. not where the print is) the drum.

3) Small ‘toner’ beads (insulators) are attracted to the drum

by an induced charge separation and stick to the drum

4) A second corotron, below the paper attracts the toner onto

the paper, where it sticks electrostatically

5) The toner and paper pass through a heat sealer where the

small plastic beads are melted onto the paper and the paper is

discharged.

3) Why is a Van de Graff generator less dangerous than an

electrical outlet box in your home?

The Van de Graff generator (although it develops a large

charge) discharges for only a fraction of a second. The wall

outlet, although at a much smaller voltage (i.e. amount of

charge) discharges continually and causes muscle tissue to

contract and remain contracted

4) Explain the process by which a lightning strike is generated.

Specifically state the type charging involved and electron

motion at each step. Include appropriate diagrams.

In the summer, there is greater heating of air just above the

ground than higher up in the atmosphere. The hotter air rises

in updrafts which tunnel up through the cooler air. This

results in frictional charging between water molecules in

different states (solid, grommel - semisolid, liquid). The

bottom part of the cloud becomes negatively charged as a

result, the middle part positively charged and the upper part

of large clouds negatively charged. In turn, the negative

bottom of the cloud repels electrons in the ground, causing an

induced positive charge on the ground. The higher pointed

objects on the ground then build up step leaders towards the

clouds (areas of positively charged pockets of air) and the

cloud negatively charged step-leaders which move towards the

ground. The lightning becomes luminous when the two

leaders meet somewhere in between and electrons move to the

ground. There is a further ground discharge as the electrons

move out laterally through the positive ground, neutralizing

the ground charge.



5) Jan 2001 - In a physics demonstration, a student inflates a

balloon by blowing into it. The end of the balloon is then tied.

The balloon is rubbed with fur and develops an electrostatic

charge. The balloon is placed against the ceiling and released.

It remains "stuck" to the ceiling.

The teacher then presents the following challenges to the

students:

-explain how the balloon received the electrostatic charge

-explain why the balloon is attracted to the ceiling

-provide a procedure that would determine if the charge on the

balloon is positive or negative. Include a list of any additional

equipment needed.

-provide a procedure that could be used to determine if there is

a relationship between the amount of rubbing and the amount

of charge developed on an inflated balloon. Include a list of

any additional equipment needed.

Using concepts from Physics 30, provide a response to each of

the teacher's challenges.

Marks will be awarded for the physics used to solve this

problem and for the effective communication of your

response.

The balloon becomes negatively charged through frictional

charging when the balloon attracts electrons away from the

fur. The charge on the balloon could be confirmed with an

electroscope of known negative charge. If the balloon is

negatively charged, the leaves of the electroscope will

separate further as the balloon is brought close to, but does

not touch the ball of the electroscope.

The negative balloon then induces a charge separation in the

wall by repelling electrons within the molecules of the wall.

The near surface of the wall gains a positive charge and the

far surface of the wall a negative charge. Since the near

surface is closer to the balloon this electrical attraction is

greater than the repulsion of the two negative surfaces.

To determine if there is a relationship between the amount of

rubbing and the amount of charge developed on the balloon:

Manipulated variable: number of rubs

Responding variable: separation of electroscope leaves

Control variable: distance of the balloon from the

electroscope.

How to change the manipulated variable: Do a series of trials

in which the number of rubs of the balloon is steadily

increased over the trials.

How to measure the responding variable: Measure the angle

of separation between the leaves of the electroscope. A

greater angle indicates a greater electrical force. (diagram)

Worksheet 3: Coulombic (Electrical) Forces

1) Two students are sitting 1.50 m apart. One student has a

mass of 70.0 kg and the other has a mass of 52.0 kg. What

is the gravitational force between them? (1.08 x 10-7 N)

1 2g 2

27

g 2

7

g

Gm mF =

R

(6.67 x 10 )(70)(52)F =

(1.50)

F = 1.08 x 10 N

−

−

2) What gravitational force does the moon produce on the

earth if the centers of the Earth and moon are 3.88 x 108 m

apart and the moon has a mass of 7.34 x 1022 kg?

(1.94 x 1020 N)

1 2g 2

27 22 24

g 8 2

20

g

Gm mF =

R

(6.67 x 10 )(7.34 x 10 )(5.98 x 10 )F =

(3.88 x 10 )

F = 1.94 x 10 N

−

3) Calculate the electric force between two point charges

of 4.00 µC and 3.00 µC when they are 2.00 cm apart.

(2.70 x 102 N)

1 2

2

9 6 6

2

2

kq qF =

R

(8.99 x 10 )(4.0 x 10 )(3.0 x 10 )F =

(0.02)

F = 2.70 x 10 N

e

e

e

− −

4) Two points of equal charge produce an electric force on

each other of 3.40 x 10-2 N when placed 1.00 x 10-1 m apart.

What is the charge on each point? (1.94 x 10-7C)

2

e1

2 1 2

1 9

7

1

F Rq =

k

3.4 x 10 (1 x 10 )q =

8.99 x 10

q = 1.94 x 10 N

− −

−



5) Two point charged objects produce an electric force on

each other of 4.5 x 10-3 N. What is the electric force if the

charge on both objects triple and the distance between them

doubles? (1.0 x 10-2 N)

1 2

2

new

1 2old2

new old

3

new

2

new

k3q 3q

F (2R) =

kq qF

R

9F = F

4

9F = (4.5 x 10 )

4

F = 1.0 x 10 N

−

−

6) Three point charged objects are placed in a line as shown in

the diagram.

Calculate the magnitude of the net electric force on the

center charge due to the other two charges. (1.1 x 10-1 N)

The net electric force will be the vector sum of the two forces

acting on the charge 3. Since the other two charges are both

positive, these forces will be pushing in opposite directions as

like charges repel

1 2e(13)

2

9 6 2

e(13)2

e(13)

e(23)

e(net) e(13) e(23)

e(net)

1

new

kq qF =

R

8.99 x 10 (2.0 x 10 )F =

(0.40)

F = 0.22475 N East

F = 0.337125 N West

F = F + -F

F = 0.22475 + -0.337125

F = 1.1 x 10 N West

−

−

�

�

�

�

� � �

�

7) Two small spheres have the same mass and volume. One of

the spheres has a charge of 4.00 µC and the other sphere has a

charge of -1.00 µC. If these two spheres are brought into brief

contact with each other and then separated to a distance of

2.00 x 10-1 m, what is the electric force between them at this

distance? (5.0 x 10-1 N)

When the two spheres touch, the total charge is shared

between the spheres

6 6

1 2

6

1 2e

2

9 6 2

e1 2

1e

+ Q 4 x 10 + -1 x 10 =

2 2

= 1.5 x 10 C each

kq qF =

R

8.99 x 10 (1.5 x 10 )F =

(2 x 10 )

F = 5.1 x 10 N repulsive

Q− −

−

−

−

−

�

�

�

8) Three point charged objects are placed at the corner of

a right-angle triangle as shown in the diagram.

Calculate the magnitude of the net electric force on the charge

marked with the x due to the other two charges. (5.0 x 10-1 N)

When charges 1 and 2 are at right angles to charge3, the net

force can be found using a Pythagorean sum of the vector

forces acting on 3. Since all the charges are positive and like

charges repel, F23 will be northward and F13 will be to the

East.



1 2e(13)

2

9 -6 -6

e(13)2

e(13)

e(23)

2 2

e(13) e(23)net

2 2

net

net

e(1

kq qF =

R

8.99 x 10 (3.0 x 10 )(4.0 x 10 )F =

(0.60)

F = 0.29966... N

F = 0.39955... N

F = F + F

F = (0.29966...) + (0.39955...)

F = 0.50 N

F = tanϑ −

�

�

�

�

� �

�

23)

e(13)

1

o

1 oe

F

0.39955... = tan

0.29966...

= 53.1 N of E

F = 5.0 x 10 N @ 53.1 N of E

ϑ

ϑ

−

−

�

�

9) Two small spheres, each with a mass of 2.00 x 10-5 kg,

are placed 3.50 x 10-1 m apart. One sphere has a charge of

-2.00 µC and is fixed in position. The other sphere has a

charge of -3.00 µC but is free to move without friction.

What is the initial acceleration due to the electric force on

the sphere that is free to move? (2.20 x 104 m/s2)

1 2e

2

9 -6 -6

e2

e

kq qF =

R

8.99 x 10 (-2.0 x 10 )(-3.0 x 10 )F =

(0.35)

F = 0.4403... N repulsive

�

�

�

e net

e

5

4

2

Since the electric force is the only force on

the object, it is the net force.

F = F

Fa =

m

0.4403...a =

2.0 x 10

m a = 2.20 x 10 away from the other charge

s

−

� �

�

10) The drawing shows three point charges fixed in place.

The charge at the coordinate origin has a value of q1 = +8.00 µC;

the other two have identical magnitudes. but opposite signs:

q2 = -5.00 µC and q3 = +5.00 µC.

a) Determine the net force (magnitude and direction) exerted

on q1 by the other two charges.

b) If q1 had a mass of 1.50 g and it was free to move, what

would be its acceleration?

e(21) e(31)

The net force will be the vector sum of the two forces

F and F . Since these vectors are at non-right

angles, the x and y components need to be added

separately to determine the net force.

F

� �

�

1 2e(21)

2

9 -6 -6

e(21)2

oe(21)

oe(31)

kq q =

R

8.99 x 10 (8.0 x 10 )(5.0 x 10 )F =

(1.3)

F = 0.21278... N at 23 N of E

direction determined as unlike charges attract

F = 0.21278... N at 23 N of W

direction d

�

�

�

etermined as like charges repel



e(21)

e(31)

e(net)

e(net)

F ( 0.21278 cos(23), +21278 sin(23))

+ F ( 0.21278 cos(23), +21278 sin(23))

F ( 0 , 0.166280 .... )

F 0.166 N North

+

−

=

�

�

�

�

11) Suppose that we place three small, equally charged metal

spheres A, B and C so that A is 1.2 cm west of B and C is

1.0 cm south of B. It is known that C exerts a force of 4.0 µN

on B.

a) What force does A exert on B? (FAB = 2.8

µN)

b) What is the magnitude and direction of the net force on B?

(4.9 µN at 35o E of N)

In this question, all the forces are repulsive since like charges

repel.

e(AB) e(CB)

1 2

2e(AB)

e(CB) 1

The net force will be the vector sum of the two forces

F and F . Since these vectors are at right

angles, a Pythagorean sum can be used to find the net force.

kq q

F (0.12) =

kqF

� �

�

�

2

22

2

26 6

e(AB) e(CB) e(CB)2

(0.1) =

q (0.12)

(0.1)

(0.1)F = F = 0.69 x 10 N F 2.8 x 10 N

(0.12)

− −=� � �

2 2

e(AB) e(CB)net

6 2 6 2

net

6

net

e(CB)1

e(AB)

61

6

o

6 o

net

F = F + F

F = (2.8 x 10 ) + (4.0 x 10 N)

F = 4.9 x 10 N

F = tan

F

4.0 x 10 = tan

2.8 x 10

= 55 N of E

F = 4.9 x 10 N @ 55 N of E

ϑ

ϑ

ϑ

− −

−

−

−−

−

−

� �

�

�

net

net

2

Since F is the only force, it is also the net force

F = F

Fa =

m

0.166a =

0.0015

ma = 111

s

e

e



12) Two equally charged identical conducting spheres A and B

repel each other with a force of 0.20 µN. Another identical

uncharged sphere, C, is touched to A and then move next to

but not touching B

a) What is the electric force on A now? (0.15 µN (left or west)

b) What is the net electric force with direction on C if it is now

moved halfway between A and B? (F net = 0.20 µN towards A)

1 2

2new

old 1

When the two conductors are touched, A and C

will share the charge equally between the two

spheres

+ + 0 = =

2 2 2

The charge where B was located will now be

3q + q =

2 2

1 3k q q

2 2F

= kq qF

A CQ Q q q

q

R

�

�

2

2

new old

3 =

4

3 3F = F = (0.20 N) = 0.15 N

4 4

The direction will be West since like charges repel

R

µ µ� �

AC BC

1 2 1 2

2 2

AC BC

old old1 2 1 2

2 2

AC ol

If C is located half way between B and A, the net

force will be the linear vector sum of F and F

1 3 1k q q k q q

2 2 2

1 1

F F2 2 = = 1 and = = 2

kq q kq qF F

F = 1F

R R

R R

� �

� �

� �

� �

d

BC old

net AC BC

net

net

0.20 N East as like charges repel

and F = 2F 0.40 N West as like charges repel

F = F + F

F = +0.20 N + -0.40 N

F = 0.20 N West

µ

µ

µ µ

µ

=

=� �

� � �

�

�

13) Three equally charged metal spheres are located as shown

The electric force exerted by A on B is 2.98 µN

a) What electric force does C exert on B? (11.9 µN)

b) What is the magnitude and direction of the net force on B?

(12.3 µN at 14° E of N)



2

1

2

CB

2AB 1

2

6CB AB

CB AB

kq

1

F 2 = = 4

kqF

F = 4F 4(2.98 x 10 )= 11.9 N North

as like charges repel

The net force can be found from a Pythagorean

sum of the forces F and F since the charges a

R

R

µ−

=

�

�

� �

� �

2 2

net AB CB

6 2 6 2net

5net

e(CB)1

e(AB)

61

6

o

net

re

at right angles to B

F = F + F

F = (2.98 x 10 ) + (11.9 x 10 )

F = 1.23 x 10 N

F = tan

F

11.9 x 10 = tan

9.98 x 10

= 76 N of E

F = 1.23 x 1

ϑ

ϑ

ϑ

− −

−

−

−−

−

� � �

�

�

�

�

�

5 o0 N @ 76 N of E−

14) A student performed an experiment that verified Coulomb's

Law of Electrostatics by measuring the repulsion between two

charged spheres, A and B, as a function of the separation of the

spheres. The spheres were identical in size and mass. The

measurements are shown in the table of value sand plotted on

the graph below. (Jan 99)

a) Show that the results verify Coulomb's Law by manipulating

the data and providing a new table of values that, when plotted,

will produce a straight-line graph.

b) Plot the new data with the responding variable on the vertical

axis.

c) Calculate the slope of your graph. (7.9 x 10-3 Nm2)

d) Using the slope value, or another suitable averaging techniques

determine the charge on sphere B if the charge on sphere A is

3.08 x 10-7 C. (2.9 x 10-6 C)

e) Determine the magnitude of the force between spheres A and B

when they are at a distance of 2.00 m apart. Use the hypothetical

value of 3.00 x 10-6 C for the charge on sphere B if you were

unable to determine the actual value. (2.0 x 10-3 N)



L1 = separation (m)

L2 = Force (N)

L3 = 1/separation2 (1/m2)

L3 1/separation2 (1/m2)

1.0 x 102

59

25

6.3

2.8

LinReg(ax+b) L1, L2, Y1

a = 7.9 x 10-3 Nm2

b = 2.0 x 10-3 N

r2 = 0.99974…

min max scl

Window x[ -6.944… 109.72… 20 ]

y[ -0.1085… 0.92056 0.1 ]

By Coulomb’s Law

1 2

e 2

2

e 1 2

2

1

2 9 7

6

2

kq qF = , therefore

R

F R = slope = kq q

slope and hence q =

kq

0.0079299...q =

(8.99 x 10 )(3.08 x 10 )

q = 2.9 x 10 C

−

−

1 2

e 2

9 7 6

e 2

3

e

kq qF =

R

(8.99 x 10 )(3.08 x 10 )(2.9 x 10 )F =

2.00

F = = 2.0 x 10 N

− −

−

Worksheet 4: Electric Fields

1) What is an electric field?

An electric field is the area around a charged particle in

which the charged particle can exert a force on other charged

particles.

2) What are the similarities between an electrical field and a

gravitational field? What are the differences between these

two types of fields?

Similarities: Both are fields that occur at a distance, and both

are inversely related to the square of the distance between the

centers of the objects involved.

Differences: Gravity is directly related to the product of two

masses and electrical is directly related to the product of two

charges. Where gravity can only be attractive, electrical

forces are both attractive (unlike charges) and repulsive (like

charges),

3) Draw the shape of the following fields:

a) around a positive point charge.

b) between a positive and a negative point charge

c) between a vertical positive and a vertical negative

plate.

d) inside a negatively charged, circular conductor.

a) b)

c) d)



4) What are the two methods to calculate the strength of an

electric field?

1

2

e

2

kqE = using the central charge which

R

creates the field

FE = using the force felt by a charge

q

which comes into the field

��

��

Worksheet 5: Electric Field Intensity

1) What is the electric field strength 7.50 x 10-1 m from a

8.00 µC charged object? (1.28 x 105 N/C)

1

2

9 6

1 2

5

kqE =

R

(8.99 x 10 )(8.0 x 10 )E =

(7.5 x 10 )

NE = 1.28 x 10

C

−

−

��

��

��

2) Calculate the gravitational field strength on the surface of

Mars. Mars has a radius of 3.43 x 106 m and a mass of

6.37 x 1023 kg. (3.61 N/kg)

1

g 2

11 23

g 6 2

g

Gma =

R

(6.67 x 10 )(6.378 x 10 )a =

(3.43 x 10 )

Na = 3.61

kg

−

3) At a point a short distance from a 4.60 x 10-6 C charged

object, there is an electric field strength of 2.75 x 105 N/C.

What is the distance to the charged object producing this

field? (3.88 x 10-1 m)

1

9 6

5

kqR =

E

8.99 x 10 (4.6 x 10 )R =

2.75 x 10

R = 0.388 m

−

��

4) If an alpha particle experiences an electric force of

0.250 N at a point in space, what electric force would a

proton experience at the same point? (0.125 N)

-19

1

2

-19

1

2

p

p

p

kq (1.60 x 10 )F R=

kq (3.20 x 10 )F

R

1F = F

2

1F = (0.250)

2

F = 0.125 N

p

α

α

5) What is the initial acceleration on an alpha particle when it

is placed at a point in space where the electric field strength is

7.60 x 104 N/C? (3.66 x 1012 m/s2)

e 2

4 19

e

14

e

e net

e

14

27

12

2

F = E q

F = (7.60 x 10 )(3.2 x 10 )

F =2.43.. x 10 N

As the electric force is the only force

present, it is the net force F = F

Fa =

m

2.43.. x 10a =

6.367 x 10

ma = 3.65 x 10

s

−

−

−

−

��

6) Calculate the electric field strength mid-way between a



4.50 µC charged object and a -4.50 µC charged object if the

two charged objects are 5.00 x 10-1 m apart. (1.29 x 106 N/C)

11

2

9 6

1

The net electric field will be the vector sum

of the field from 1 and the field from 2, where

the directions of the fields are defined by a

positive test charge

kqE =

R

8.99 x 10 (4.5 x 10 )E =

(2.5

−

��

��

1 2

51

52

net 1 2

5net

6net

x 10 )

NE = 6.47... x 10 right (see diagram)

C

NE = 6.47... x 10 right (see diagram)

C

E = E + E

E = 2(6.47... x 10 )

NE = 1.29 x 10 right

C

−

��

��

��

��

��

7) Calculate the electric field strength mid-way between a

3.0 µC charged object and a 6.0 µC charged object if the

objects are 8.0 x 10-1 m apart. (1.7 x 105 N/C)

The net electric field will be the vector sum

of the field from 1 and the field from 2, where


positive test charge

11

2

9 6

11 2

51

52

net 1 2

5 5net

net

kqE =

R

8.99 x 10 (3.0 x 10 )E =

(4.0 x 10 )

NE = 1.685 x 10 East (see diagram)

C

NE = 3.371 x 10 West (see diagram)

C

E = E + - E

E = 1.685 x 10 + -3.371 x 10

E = 1.7 x

−

−

��

��

��

��

��

��

��

5 N10 West

C

8) Calculate the magnitude and direction of the electric field at

the centre of a square with 10.0 cm sides if the corners taken

in counter-lockwise rotation from top right, have charges of

+1.00 µC, +2.00 µC, +3.00 µC and +4.00 µC. The square is

level with the horizontal. (8.99 x 104 N/C toward the top of the

square)

2 2

2 2

The distance from each of the charges to the

point can be deduced from the Pythagorean

relationship.

R = x + y

R = 0.05 + 0.05

R = 0.7071 m

Each of the charges is at a 45 angle to the point

due to the

o

sides of the triangle in the Pythagorean

relationship being equal.



Since the fields created by the three charges

are not all at right angles to the point, the

net electric field will be the vector sum

of the field from 1, 2, 3 and 4, expressed as

components in the x and

11

2

9 6

12

61

62

3

y dimensions where


positive test charge.

kqE =

R

8.99 x 10 (1.0 x 10 )E =

(0.07071)

NE = 1.798 x 10 45 S of W

C

NE = 3.596 x 10 45 S of E

C

E =

o

o

−

��

��

��

��

��

6

64

61

N 5.394 x 10 45 N of E

C

NE = 7.192 x 10 45 N of W

C

Using component techniques to resolve the net

electric field.

x y

E = (-1.798 x 10 cos 45, 1.7

o

o��

��

6

6 62

6 63

6 64

net

98 x 10 sin 45)

E = (3.596 x 10 cos 45, 3.596 x 10 sin 45)

E = (5.394 x 10 cos 45, 5.394 x 10 sin 45)

+ E = (-7.192 x 10 cos 45, 7.192 x 10 sin 45)

E = ( 0 , 5.09

−��

��

��

��

6 x 10 )

6net

NE = 5.09 x 10 North

C

��

9) A charged sphere having 2.50 C of excess positive charge

is located 50.0 m due north of a second sphere having 10.0 C

of excess positive charge. What is the net electrical field

intensity at the point(s) 30.0 m from the first and 40.0 m from

the second sphere? (6.15 x 107 N/C 60.8o E of N)

-1

-1

Since this is a 3, 4, 5 triangle, it must have

a right angle as indicated.

o= tan = 90 -

a

30= tan = 90 - 36.869

40

= 36.869 = 53.13o o

ϑ φ ϑ

ϑ φ

ϑ φ

Since the fields created by the two charges

are in a coordinate plane, the

net electric field will be the vector sum

of the field from 1 and 2, expressed as

components in the x and y dimensions where


positive test charge.



1A

2

9

A2

7A

7B

kqE =

R

8.99 x 10 (2.5)E =

(30)

NE = 2.497 x 10 36.869 S of E

C

NE = 5.618... x 10 53.13 N of E

C

Using component techniques to resolve the net

electric field.

x

o

o

��

��

��

��

7 7A

7 7B

7net

y

E = (2.49... x 10 cos36.869, -2.49... x 10 sin 36.869)

+ E = (5.618... x 10 cos53.13, 5.618... x 10 cos53.13)

E = ( 5.37... x 10 ,

��

��

��

7

net(y)2 21

net net(x) net(y)

net(x)

7

7 2 7 2 1net

7

7net

2.99... x 10 )

EE E + E = tan

E

2.99... x 10E (5.37... x 10 ) + (2.99... x 10 ) = tan

5.37... x 10

NE = 6.15 x 10 = 29.1

C

ϑ

ϑ

ϑ

−

−

=

=

��

��

��

��

��

7 0net

NE = 6.15 x 10 at 29.1 N of E

C

o

��

10) June 89. All three objects are positively charged

a) Given that the charges on A and B are each of magnitude

3.00 x 10-6 C, and that the charge on C is of magnitude

4.00 x 10-6 C, determine the magnitude and direction of the

net force acting on object A. Illustrate the answer to part with

an appropriate sketch. (37.5 N at 55.1o S of W)

b) If the mass of A is 3.2 x 10-3 kg, what is the initial

acceleration of A? (1.2 m/s2 at 55.1o S of W)

d) Use a vector sketch to draw in the approximate direction of

the electric field intensity at point P which is half way between

B and C.

BA

CA

1 2BA

2

9 6

BA

The net force will be the vector sum of F

and F where the forces are at right angles

and hence the net force can be determined using

a Pythagorean sum.

kq qF =

R

8.99 x 10 (3.0 x 10 )(3.0 x 1F =

−

�

�

�

�

6

2

BA

CA

0 )

(0.06)

F = 22.475 N West

F = 29.96 N South

Where the directions of the forces are determined

from the law of charges that like charges repel.

−

�

�



2 2 CA1eNet BA CA

BA

2 2 1eNet

eNet

0eNet

FF F + F = tan

F

29.96F (22.475) + (29.96) = tan

22.475

F = 37.5 N = 53.1

F = 37.5 N at 53.1 S of W

o

ϑ

ϑ

ϑ

−

−

=

=

�

� � �

�

�

�

�

e net

e

3

4

2

As the electric force is the only force operating, it is also

the net force F = F

Fa =

m

37.5a =

3.2 x 10

ma = 1.2 x 10 at 53.1 S of W

s

o

−

Worksheet 6: Electrical Potential and Field Potential:

Point Charges

1) What is the potential at a distance of 6.0 cm from a

2.5 µC charge? (3.7 x 105 V)

1

9 6

5

kqV =

R

8.99 x 10 (2.5 x 10 )V =

0.06

V = 3.7 x 10 V

−

2) Three charges are along a line as shown below:

Find the potential at point P. (-2.0 x 105 V)

1

1

9 6

1

5

1

The voltage at P will be the scalar sum of

the three voltages.

kqV =

R

8.99 x 10 (1.0 x 10 )V =

0.075

V = 1.2 x 10 V

−

5

2

5

3

net 1 2 3

5 5 5

net

5

net

V = -1.5 x 10 V

V = -1.6 x 10 V

V = V + V + V

V = 1.2 x 10 + -1.5 x 10 + -1.6 x 10

V = -2.0 x 10 V

3) Three charges are at the corners of a rectangle as shown

below:

Find the potential at point P. (4.4 x 105 V)

2 2

2 2

The distance from charge 2 to the

point can be deduced from the Pythagorean

relationship.

R = x + y

R = .060 + 0.080

R = 0.10 m



1

1

9 6

1

5

1

5

2

5

2

net 1 2 3

5

net


the three voltages.

kqV =

R

8.99 x 10 (1.0 x 10 )V =

0.080

V = 1.12 x 10 V

V = -2.697 x 10 V

V = 5.993 x 10 V

V = V + V + V

V = 1.12 x 10 + -2.697

−

5 5

5

net

x 10 + 5.993 x 10

V = 4.4 x 10 V

4) A -3.5 µC charge is 4.0 cm to the left of a 5.0 µC charge.

What is the field potential at a point 1.0 cm to the right of the

-3.5 µC charge? (-1.6 x 106 V)

1

1

9 6

1

6

1

6

2

net 1 2

6 6

net

net


the two voltages.

kqV =

R

8.99 x 10 (-3.5 x 10 )V =

0.010

V = -3.1465 x 10 V

V = 1.123 x 10 V

V = V + V

V = -3.1465 x 10 + 1.123 x 10

V = -1.6

−

6x 10 V

5) What is the field potential at a distance of 1.00 x 10-10 m

from a proton? (14.4 V)

1

1

9 19

1 10

1

kqV =

R

8.99 x 10 (1.60 x 10 )V =

1.0 x 10

V = 14.4 V

−

−

6) What is the electrical potential energy of an electron at a

distance of 1.00 x 10-10 m from a proton? (-2.30 x 10-18 J ,

negative indicates attraction)

1 2

el

9 19 19

el 10

18

el

kq qE =

R

8.99 x 10 (1.60 x 10 )(-1.60 x 10 )E =

1.0 x 10

E = 2.30 x 10 J

− −

−

−−

7) What is the potential energy of a proton half way between

two alpha particles separated by 1.00 x 10-14 m?

(1.84 x 10-13 J)

1 2

el1

9 19 19

el1 15

14

el1

14

el2

elnet e

The electrical potential energy at P will be

the scalar sum of the two energies.

kq qE =

R

8.99 x 10 (3.20 x 10 )(1.60 x 10 )E =

5.00 x 10

E = 9.20 x 10 J

E = 9.20 x 10 J

E = E

− −

−

−

−

l1 el2

14

elnet

13

elnet

+ E

E = 2(9.20 x 10 )

E = 1.84 x 10 J

−

−



8) How much does the potential energy of a 0.030 µC charge

in the field of a 5.0 µC charged object change when it is

moved from R = 45 cm away to R = 15 cm? (6.0 x 10-3 J)

el el(final) el(initial)

1 2 1 2

el

f

el 1 2

f i

9 8 6

el

3

el

∆E = E - E

kq q kq q∆E = -

R R

1 1∆E = kq q -

R R

∆E = 8.99 x 10 (3.0 x 10 )(5.0 x 10 )

1 1 -

0.15 0.45

∆E = 6.0 x 10 J

i

− −

−

9) 4.40 x 10-5 J of energy is used in moving a 3.00 µC charge

at a constant speed from point A to point B. If A and B are

2.4 cm apart, what is the potential difference between A and

B? (14.7 V)

2

5

6

EV =

q

4.40 x 10V =

3.00 x 10

V = 14.7 V

−

−

Λ∆

∆

∆

Worksheet 7: Electrical Potential and Field Potential:

Plates

1) Two parallel plates are connected to a 12.0 V battery. If

the plates are 9.00 x 10-2 m apart, what is the electric field

strength between them? (1.33 x 102 V/m)

2

VE =

d

12E =

0.0900

VE = 1.33 x 10

m

��

��

��

2) The electric field between two parallel plates is

5.0 x 103 V/m. If the potential difference between the

plates is 2.0 x 102 V, how far apart are the plates?

(4.0 x 10-2 m)

3

2

Vd =

E

200d =

5.0 x 10

d = 4.0 x 10 m−

��

3) What is the value of the potential energy of an alpha

particle in the electric field between parallel plates if it is right

next to the positive plate at 500.0 V? (1.60 x 10-16)

el 2

19

el

16

el

E = Vq

E = 500(3.2 x 10 )

E = 1.6 x 10 J

−

−

4) If an electron is 2.00 cm away from the negative plate in a

1.00 x 103 N/C electric field between parallel plates which are

8.00 cm apart, what is the value of the electron’s potential

energy? (9.60 x 10-18 J)

3

V = E d

V = 1.0 x 10 (0.080)

V = 80 V

��

An electron that is 2.00 cm away from the negative plate still

retains ¾ of its potential energy in the 8.00 cm field between

the plates.

el 2

19

el

18

el

E = Vq

E = 60(1.6 x 10 )

E = 9.60 x 10 J

−

−



5) What is the field intensity between two parallel plates if an

electron is halfway between the plates? The plates have a

potential of 750 V and are 10.0 cm apart. (7.50 x 103 N/C)

3

E is constant between parallel plates of

opposite charge

VE =

d

750E =

0.10

VE = 7.50 x 10

m

��

��

��

��

6) What is the potential difference between two parallel plates

5.00 cm apart that produces an electric force of 3.50 x 10-13 N

on an object containing a charge of 2.00 x 10-16 C when it is

placed between the plates? (8.75 x 101 V)

e

2

13

16

FV = d

q

3.5 x 10V = (0.050)

2.0 x 10

V = 87.5 V

−

−

7) The electric field strength between two parallel plates

is 9.3 x 102 V/m when the plates are 7.0 cm apart. What would

the electric field strength be if the plates were 5.0 cm apart?

(1.8 x 103 V/m)

1 2

el1 k2

k

k

k(new)

new

k(old)old

k(new)

new old

k(old)

17

2

new 17

Energy is conserved as electrical potential

is converted to kinetic energy.

E = E

E = E

Vq = E

E V =

q

E

V q=

EV

q

EV = V

E

9.00 x 10V = (4.2 x 10 )

3.00 x 10

V

−

−

Σ Σ

∴

3

new = 1.26 x 10 V

8) A helium nucleus, with a mass of 6.65 x 10-27 kg, is

inserted between two charged parallel plates. The potential

difference between the plates is 1.5 x 103 V, and the distance

between the plates is 3.5 cm.

What time is required for the helium nucleus to reach the

negative plate?

net

e

2

2

3 19

27

12

2

2

i

F1) a = since F is the only significant force

m

Fa =

m

a =

m

V qa =

d m

1.5 x 10 (3.2 x 10 )a =

(0.035)(6.65 x 10 )

ma = 2.06 x 10

s

12) d = v t + at since v = 0

2

2dt =

a

2(0.035)t =

2.06

e

i

E q

−

−

��

12

7

x 10

t = 1.84 x 10 s−

9) What is the potential difference between two parallel

charged plates that are 7.50 cm apart if a force of

5.30 x10-14 N is needed to move an alpha particle from the

negative plate to the positive plate?(1.24 x 104 V)

e

2

14

16

4

FV = d

q

5.3 x 10V = (0.075)

3.2 x 10

V = 1.24 x 10 V

−

−

10) An alpha particle is placed between two horizontal parallel

charged plates that are 2.00 cm apart. The potential difference

between the plates is 12.0 V.

a) What is the electric force acting on the alpha particle?

(1.92 x 10-16 N)

b) What is the gravitational force acting on the alpha particle?

(6.52 x 10-26 N)

c) Assuming that the electric force and the gravitational force

are acting in opposite directions, what is the net force acting

on the alpha particle? (1.92 x 10-16 N)



d) What is the acceleration of the alpha particle?

(2.88 x 1010 m/s2)

e) What potential difference would be required between the

plates in order that the alpha particle becomes suspended?

(4.09 x 10-9 V)

a)

e

e 2

e 2

19

e

16

e

Both F and E are constant between parallel

plates of opposite charge

F = E q

VF = q

d

12F = (3.2 x 10 )

0.020

F = 1.92 x 10 N

−

−

��

��

b)

g

27

g

26

g

F = mg

F = 6.67 x 10 (9.81)

F = 6.52 x 10 N

−

−

c)

net el g

16 26net

16net

F = F + -F

F = 1.92 x 10 + -6.52 x 10

F = 1.92 x 10 N

− −

−

� � �

�

�

d)

net

16

27

10

2

Fa =

m

1.92 x 10a =

6.67 x 10

ma = 2.88 x 10

s

−

−

�

�

�

�

e)

el g

2

2

27

19

9

If a particle were suspended between the

plates then forces would be balanced.

F = F

Vq = mg

d

mgdV =

q

6.67 x 10 (9.81)(0.020)V =

3.2 x 10

V = 4.09 x 10 V

−

−

−

� �

Worksheet 8: Work and Energy in Electric Fields

1) A proton is released 2.0 x 10-11 m from the center of a

6.4 x 10-18 C charged sphere. What is the speed of this proton

when it is 0.50 m from this center? (7.4 x 105 m/s)

1 2

el1 k2 el12

21 2 1 2

2

1 2

1 2

2

1 2

9 19

2

Energy is conserved as some of the electrical

potential energy is converted to kinetic energy

E = E

E = E + E

kq q kq q1 = mv +

R 2 R

2kq q 1 1v = -

m R R

2(8.99 x 10 )(1.6 x 10v =

−

Σ Σ

18

27 11

5

2

)(6.4 x 10 ) 1 1 -

1.67 x 10 2 x 10 0.50

mv = 7.4 x 10

s

−

− −

2) The centers of two alpha particles are held 2.5 x 10-12 m

apart when they are released. Calculate the speed of each

alpha particle when they 0.75 m apart. (2.4x 104 m/s)



1 2

el1 k2 el12

21 2 1 2

2

1 2

1 2

2

1 2



of the two alpha particles

E = E

E = 2E + E

kq q kq q1 = (2) mv +

R 2 R

kq q 1 1v = -

m R R

Σ Σ

9 19 2

2 27 12

5

2

2(8.99 x 10 )(3.2 x 10 ) 1 1v = -

6.67 x 10 2.5 x 10 0.75

mv = 2.3 x 10

s

−

− −

3) An alpha particle gains 1.50 x 10-15 J of kinetic energy.

Through what potential difference was it accelerated?

(4.69 x 103 V)

1 2

el1 k2

k2

k2

15

19

3

Energy is conserved as the electrical


E = E

E = E

Vq = E

EV =

q

1.5 x 10V =

3.2 x 10

V = 4.69 x 10 V

−

−

Σ Σ

4) A proton is accelerated by a potential difference of

7.20 x 102 V. What is the change in kinetic energy of the

proton? (1.15 x 10-16 J)

1 2

el1 k2

k2

2 19

k2

16

k2



E = E

E = E

Vq = E

E = 7.2 x 10 (1.6 x 10 )

E = 1.15 x 10 J

−

−

Σ Σ

5) What maximum speed will an alpha particle reach if it

moves from rest through a potential difference of

7.50 x 103 V? (8.50 x 105 m/s)

1 2

el1 k2

2

2

3 19

27

5



E = E

E = E

mvVq =

2

2Vqv =

m

2(7.5 x 10 )(3.2 x 10 )v =

6.67 x 10

mv = 8.48 x 10

s

−

−

Σ Σ

6) A proton is placed in an electric field between two parallel

plates. If the plates are 6.0 cm apart and have a potential

difference between them of 7.5 x 101 V, how much work is

done against the electric field when the proton is moved

3.0 cm parallel to the plates? (0)

If the proton moves horizontally, there is no change in its

potential energy. Hence, there is no work done.

7) In the above question, how much work would be done

against the electric field if the proton was moved 3.0 cm

perpendicular to the plates? (6.0 x 10-18 J)

If the charge moves vertically 3.0 cm out of a total of 6.0 cm,

then it will cross half of the voltage between the plates. From

the work-energy theorem, the work done is equivalent to the

change in potential energy.



el

2

19

18

W = ∆E

W = ∆Vq

W = 37.5(1.6 x 10 )

W = 6.0 x 10 J

−

−

8) A charged particle was accelerated from rest by a potential

difference of 4.20 x 102 V. If this particle increased its kinetic

energy to 3.00 x 10-17 J, what potential difference would be

needed to increase the kinetic energy of the same particle to

9.00 x 10-17 J? (1.26 x 103 V)

1 2

el1 k2

k

k

k(new)

new

k(old)old

k(new)

new old

k(old)

17

2

new 17

Energy is conserved as electrical potential

is converted to kinetic energy.

E = E

E = E

Vq = E

E V =

q

E

V q=

EV

q

EV = V

E

9.00 x 10V = (4.2 x 10 )

3.00 x 10

V

−

−

Σ Σ

∴

3

new = 1.26 x 10 V

9) An alpha particle with an initial speed of 7.15 x 104 m/s

enters through a hole in the positive plate between two parallel

plates that are 9.00 x 10-2 m apart as shown in the diagram.

If the electric field between the plates is 1.70 x 102 V/m, what

is the speed of the alpha particle when it reaches the negative

plate? (8.11 x 104 m/s)

1 2

k1 el1 k2

2 2

1 2

2

1

2

3 -19 -27



E = E

E + E = E

Emv mv + q =

2 d 2

2 E q - mdvv =

md

2(1.7 x 10 )(3.2 x 10 ) - 6.67 x 10v =

Σ Σ

��

��

4 2

27

4

(0.090)(7.15 x x 10 )

6.67 x 10 (0.090)

mv = 8.11 x 10

s

−

10) An electron with a speed of 5.0 x 105 m/s enters through a

hole in the positive plate and collides with the negative plate at

a speed of 1.0 x 105 m/s. What is the potential difference

between the plates? (6.8 x 10-1 V)

1 2

k1 k2 el2

2 2

1 2

2 2

1 2

-31 5 2 5 2

Energy is conserved as some of the kinetic

energy is converted to electrical potential energy

E = E

E = E + E

mv mv = + Vq

2 2

m(v - v )V =

2q

(9.11 x 10 )[(5.0 x10 ) - (1.0 x10 ) ]V =

2(1.

Σ Σ

1960 x 10 )

V = 0.68 V

−

Worksheet 9: Millikan’s Oil Drop Experiment

1) An oil drop weighs 3.84 x 10-15 N. If it is suspended

between two horizontal parallel plates where the electric field

strength is 1.20 x 104 N/C, what is the magnitude of the charge

on the oil drop? (3.20 x 10-19 C)



e g

2

2

15

2 4

19

2

If the oil droplet is held stationary, then

the upward electric force must balance with

the downward gravitational force.

F = F

E q = mg

mgq =

E

3.84 x 10q =

1.2 x 10

q = 3.20 x 10 C

−

−

��

��

2) An oil drop with a mass of 4.80 x 10-16 kg is suspended

between two horizontal parallel plates that are 6.00 cm apart.

If the potential difference between the plates is 5.90 x 102 V,

how many excess electrons does the oil drop carry? (3)

194.788 x 10 C−

e g

2

2

2

16

2 2

2

If the oil droplet is held stationary, then

the upward electric force must balance with

the downward gravitational force.

F = F

E q = mg

Vq = mg

d

mgdq =

V

4.8 x 10 (9.81)(0.06)q =

5.9 x 10

q = 4.788

−

��

19

19

19

-

2

x 10 C

1 e4.788 x 10 C x

1.6 x 10 C

q = 3e

−

−

−

−

3) An oil drop with a mass of 7.20 x 10-16 kg is moving up at a

constant speed of 2.50 m/s between two horizontal parallel

plates on the inside of an evacuated iron pot like Millikan’s. If

the electric field strength is 2.20 x 104 V/m, what is the

magnitude of the charge on the oil drop? (3.21 x 10-19 C)

If the oil droplet is moving upward at a constant

velocity, then the upward electric force must

balance with the downward gravitational force.



e g

2

2

16

2 4

19

2

F = F

E q = mg

mgq =

E

7.2 x 10 (9.81)q =

2.20 x 10

q = 3.21 x 10 C

−

−

��

��

4) An oil drop whose mass is 3.50 x 10-15 kg accelerates

downward at a rate of 2.50 m/s2 when placed between two

horizontal parallel plates that are 1.00 cm apart. Assuming the

oil drop is negative and the top plate positive, how many

excess electrons does the oil drop carry if the potential

difference between the plates is 5.38 x 102 V? (3)

net e g

2

2

15

2

If the oil droplet is accelerating downward

velocity, then the upward electric force must

be less than the downward gravitational force.

F = -F + F

ma = - E q + mg

md(a - g)q =

-V

3.5 x 10 (0.0q =

−

��

2

19

2

19

19

-

2

1)(2.5 - 9.81)

-5.38 x 10

q = 4.755 x 10 C

1 e4.755 x 10 C x

1.6 x 10 C

q = 3e

−

−

−

−

5) During a Millikan oil drop experiment, a student records the

weight of five different oil drops. A record is also made of the

Potential Difference necessary to hold each drop stationary

between two horizontal parallel plates. (slope = 2/4 x 1014 N)

Mass (10-16 kg) Potential Difference (103V)

8.8 2.03

6.3 1.58

4.5 1.15

3.5 0.81

1.3 0.27

a) Draw a graph showing voltage as a function of weight.

b) What does the slope of the graph represent?

c) If the distance between the plates was a constant 0.40 m,

what is implied about the charge on each of the drops?

(q = 1.6 x 10-18 C)

L1 = Mass (10-16 kg)

L2 = Voltage (V)

L3 = Weight of Fg (N)

L3 Weight (N) Manipulated = Wieght (N)

8.6 x 10-18 Responding = Voltage (V)

6.2 x 10-18 Control = distance between plates

4.4 x 10-18

3.4 x 10-18

1.3 x 10-18

LinReg(ax+b) L3, L2, Y1

a = 2.4 x 1017 V/N

b = 15 V

r2 = 0.9903…

min max scl

Window x[ 5.395… x 10-16 9.3688… x 10-16 5 x 10-15 ]

y[ -29.2… 2329.2 200 ]

In a balanced situation like this, the upward forces will be

equal to the downward forces.



e g

2 g

2 g

g

2

g

2

If the oil droplet is stationary,

then the upward electric force must

equal the downward gravitational force.

F = F

E q = F

Vq = F

d

F dV =

q

Since slope = V/F using the equation above

dslope = and he

q

��

2

2 17

19

2

dnce q =

slope

0.40q =

2.4 x 10

q = 1.6 x 10 C−

6) An oil drop whose mass is 5.70 x 10-16 kg accelerates

upward at a rate of 2.90 m/s2 when placed between two

horizontal parallel plates that are 3.50 cm apart. If the

potential difference between the plates is 7.92 x 102 V, what is

the magnitude of the charge on the oil drop? (3.20 x 10-19 C)

net e g

2

2

2

1

2

If the oil droplet is accelerating upward,

then the upward electric force must be

greater than the downward gravitational force.

F = F + -F

ma = E q + -mg

Vma = q + -mg

d

md(a + g)q =

V

5.7 x 10q =

−

��

6

2

19

2

(0.035)(2.9 - 9.81)

2.92 x 10

q = 3.20 x 10 C−

7) In a Millikan oil drop experiment a student sprayed oil

droplets with a density of 7.8 x 102 kg/m3 between two

horizontal parallel plates that were 4.0 cm apart. The student

adjusted the potential difference between the plates to

4.6 x 103 V so that one of the drops became stationary.

The diameter of this drop was measured to be 2.4 x 10-6 m.

What was the magnitude of the charge on this oil drop?

(Vsphere = 4/3πr3) (4.8 x 10-19 C)

e g

2 g

2

2

3

2

2 6 3

2



equal than the downward gravitational force.

F = F

E q = F

Vq = mg

d

mgdV =

q

4( r )gd3q =

V

47.8 10 ( [1.2 x 10 ] )9.81(0.040)

3q =

x

ρ π

π −

��

3

19

2

4.6 x 10

q = 4.8 x 10 C−

8) An oil droplet (q = 5e-) is suspended between two parallel

charged plates (V = 175 V). If an oil droplet of the same mass

but a charge of 3e- to be suspended between the same plates,

what potential difference would be necessary (292 V)






e g

2 g

2

2

F = F

E q = F

Vq = mg

d

mgdV =

q

mgdV 53e= =

mgdV 3

5e

5V = V

3

5V = (175)

3

V = 292 V

new

old

new old

new

new

−

−

��

9) In an oil drop experiment similar to Millikan's, an oil

droplet is suspended between two parallel charged plates.

Calculate the magnitude of the charge on the oil droplet

given the following:

• radius of the oil drop 4.2 x 10-6 m

• density of the oil 7.8 x 102 kg/m3

• distance between plates 2.0 cm

• potential difference between plates 99 V

(4.80 x 10-16 C)

e g

2 g

2




F = F

E q = F

Vq = mg

d

��

2

3

2

2 6 3

2

19

2

gdq =

V

4r gd

3q =

V

47.8 x 10 ( [4.2 x 10 ] )9.81(0.020)

3q = 79

q = 4.8 x 10 C

Vρ

ρ π

π −

−

10) In Millikan's oil drop experiment an oil drop having a

mass of 6.00 x 10-17 kg is accelerating upward at a rate of

4.60 m/s2 when sprayed between two horizontal parallel

plates 5.00 cm apart. If the potential difference between

the plates is 9.00 x 101 V, what is the charge on the oil

drop? (4.80 x 10-19 C)

net e g

2

2

2

1

2

If the oil droplet is accelerating upward,

then the upward electric force must be

greater than the downward gravitational force.

F = F + -F

ma = E q + -mg

Vma = q + -mg

d

md(a + g)q =

V

6.0 x 10q =

−

��

6

19

2

(0.050)(4.6 + 9.81)

90

q = 4.80 x 10 C−

Worksheet 10: Electric Current



1) A current of 3.60 A flows for 15.3 s through a conductor.

Calculate the number of electrons that pass through a point in

the conductor in this time. (3.44 x 1020)

19

20

Q = I t

Q = (3.60)(15.3)

1Q = 55.08 C x

1.6 x 10 C

Q = 3.44 x 10 e

e−

−

−

2) How long would it take 2.0 x 1020 electrons to pass through

a point in a conductor if the current was 10.0 A?

(3.2 s)

Qt =

I

32t =

10

t = 3.2 s

3) Calculate the current through a conductor, if a charge of

5.60 C passes through a point in the conductor in 15.4 s.

(3.64 x 10-1 A)

QI =

t

5.6I =

15.4

I = 0.364 A

4) If 5.26 x 1020 alpha particles pass a point in space every

10.0 s, what is the current at this location? (16.8 A)

QI =

t

168.33I =

10

I = 16.8 A

5) If 34.0 x 10-3 A of electron current passes between the

terminals of a 1.50 V battery in 1.20 s, how much energy has

the battery used up in this time? (6.12 x 10-2 J)

2

3

2

E = Vq where Q = It

gives E = VIt

E = 1.5(34 x 10 )(1.2)

E = 6.12 x 10 J

−

−

6) If a current of 3.20 A passes through an electric circuit for

4.00 s, how many electrons have passed any given point in the

circuit? (8.00 x 1019 e)

19

19

Q = I t

Q = (3.22)(4)

1Q = 12.8 C x

1.6 x 10 C

Q = 8.00 x 10 e

e−

−

−

7) There are 7.50 x 1027 protons traveling through a region of

space every hour. What is the electric current at this point?

(3.33 x 105 A)

19

27 9

+

9

3

5

1.60 x 10 C7.5 x 10 p x = 1.20 x 10 C

1 p

QI =

t

1.20 x 10I =

3.6 x 10

I = 3.33 x 10 A

−

+



8) When an electric appliance is connected to a

1.20 x 102 V power line, there is a current through the

appliance of 18.3 A. What is the average amount of energy

given to each electron per second by the power line?

(1.92 x 10-17 J/s)

2

3

19

20

3

20

17

-

E = Vq where Q = It

gives E = VIt

E = 120(18.3)(1)

E = 2.196 x 10 J

118.3 C x

1.60 x 10 C

1.14375 x 10

E 2.196 x 10 =

1.14375 x 10

J1.92 x 10

e

e

e

e

−

−

−

−

−

=

=

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