Physics 202 - University of Virginiagalileo.phys.virginia.edu/~pqh/202_1n.pdf · Electric Charges, Forces and Fields A problem: 3 charges lie along the x-axis. Charge q1 = 1 C is

Physics 202Professor P. Q. Hung

311B, Physics Building

Physics 202 – p. 1/27

Electric Charges, Forces and Fields

http://galileo.phys.virginia.edu/~pqh/202/PHYS_202_ homepage.htm

Important informations contained in thehomepage.

Physics 202 – p. 2/27


http://galileo.phys.virginia.edu/~pqh/202/PHYS_202_ homepage.htm

Important informations contained in thehomepage.

Physics 202 – p. 2/27


A problem:3 charges lie along the x-axis. Chargeq1 = 1µC is at the origin. Charge q2 = −3µCis at x = −0.30m. Charge q3 = −4µC is atx = +0.20m. Find the net electrostatic forceacting on q1.

What is an electric charge? What is thisstrange unit µC? What is the meaning of anelectrostatic force on one charge in thepresence of two other charges? Is it a contactforce or not?

Physics 202 – p. 3/27


A problem:3 charges lie along the x-axis. Chargeq1 = 1µC is at the origin. Charge q2 = −3µCis at x = −0.30m. Charge q3 = −4µC is atx = +0.20m. Find the net electrostatic forceacting on q1.

What is an electric charge? What is thisstrange unit µC? What is the meaning of anelectrostatic force on one charge in thepresence of two other charges? Is it a contactforce or not?

Physics 202 – p. 3/27


Electric charges

Electricity comes from the Greek elektron =Amber.

Rub an amber rod and you can pick up littlepieces of papers, a phenomenon usuallyreferred to as “static electricity”.⇒ The rod becomes “charged” by rubbing.

You rub two amber rods and put them closeto each other: They repel each other.

You rub two glass rods and put them close toeach other: They repel each other.

Physics 202 – p. 4/27


Electric charges





Physics 202 – p. 4/27


Electric charges





Physics 202 – p. 4/27


Electric charges





Physics 202 – p. 4/27


Electric charges

You rub one amber rod and one glass rod andput them close to each other: They attracteach other. ⇒ Two kinds of charges!

Like charges repel while unlike chargesattract.

Benjamin Franklin: positive charge for therubbed glass rod and negative charge for therubbed amber rod. Arbitrary choice!

Physics 202 – p. 5/27


Electric charges




Physics 202 – p. 5/27


Electric charges




Physics 202 – p. 5/27


Electric charges

Electric charge is conserved! If some positiveamount of charges is produced somewhere,an equal amount of negative charge is alsoproduced.

The unit of charge is C which stands forCoulomb.

Where do these charges come from? Fromthe atom!

Physics 202 – p. 6/27


Electric charges




Physics 202 – p. 6/27


Electric charges




Physics 202 – p. 6/27


Electric charges: The Atom

Physics 202 – p. 7/27


Electric charges

Atom: object with a nucleus (made out ofprotons and neutrons) carrying a net positivecharge surrounded by electrons carrying anet negative charge of the same magnitude.Atoms are neutral.

The charge of the proton is+|e| = 1.6 × 10−19 C and that of the electron is−|e| = −1.6 × 10−19 C. That of the neutron iszero. A neutral atom has an equal number ofprotons and electrons.

Physics 202 – p. 8/27


Electric charges

Atom: object with a nucleus (made out ofprotons and neutrons) carrying a net positivecharge surrounded by electrons carrying anet negative charge of the same magnitude.Atoms are neutral.

The charge of the proton is+|e| = 1.6 × 10−19 C and that of the electron is−|e| = −1.6 × 10−19 C. That of the neutron iszero. A neutral atom has an equal number ofprotons and electrons.

Physics 202 – p. 8/27


Electric charges

Charges can be exchanged by removing oradding electrons.

A neutral atom with electrons removed ⇒ apositive ion; added ⇒ a negative ion.

Nucleus: positively-charged protons +electrically neutral particles: neutrons.Modern theory of particle physics: protonsand neutrons are made out of even smaller,electrically charged particles: the quarks. Weare all made out of charged particles!

Physics 202 – p. 9/27


Electric charges




Physics 202 – p. 9/27


Electric charges




Physics 202 – p. 9/27


How does a charged rod attract a neutral pieceof paper?

Physics 202 – p. 10/27


Insulators and Conductors

Materials with electrons loosely bound to thenuclei and which can “freely” move within ⇒conductor. “Free” electrons: conductionelectrons. Examples are metals.

Materials with electrons very tightly bound tothe nuclei and which cannot freely moveabout: insulators. Wood is an example.

In between, materials with few conductionelectrons and which have interestingproperties: semiconductors. Silicon is anexample.

Physics 202 – p. 11/27





In between, materials with few conductionelectrons and which have interestingproperties: semiconductors. Silicon is anexample.

Physics 202 – p. 11/27





In between, materials with few conductionelectrons and which have interestingproperties: semiconductors. Silicon is anexample. Physics 202 – p. 11/27


Charging a Conductor

Physics 202 – p. 12/27


Coulomb’s Law

Recall the attractive gravitational forcebetween two masses, m1 and m2:

Fg = −Gm1 m2

r2

Experimental discovery by Coulomb:Coulomb’s Law. Electrostatic force betweentwo point charges, q1 and q2:

Fe = k q1 q2

r2 (1)

k = 8.99× 109 N.m2/C2. (1) is valid not just forpoint charges but also for charge distributions

Physics 202 – p. 13/27


Coulomb’s Law

Recall the attractive gravitational forcebetween two masses, m1 and m2:

Fg = −Gm1 m2

r2

Experimental discovery by Coulomb:Coulomb’s Law. Electrostatic force betweentwo point charges, q1 and q2:

Fe = k q1 q2

r2 (1)

k = 8.99× 109 N.m2/C2. (1) is valid not just forpoint charges but also for charge distributions

Physics 202 – p. 13/27


Gravitational force vs Electrostatic force

The gravitational force is always attractivewhile the electrostatic force can be attractiveor repulsive depending on the sign of theproduct q1 q2.

The electrostatic force is attractive whenq1 q2 < 0, i.e. the two charges have oppositesigns (one positive and the other onenegative).

Fe = −k |q1| |q2|r2 (1)

Physics 202 – p. 14/27



The gravitational force is always attractivewhile the electrostatic force can be attractiveor repulsive depending on the sign of theproduct q1 q2.

The electrostatic force is attractive whenq1 q2 < 0, i.e. the two charges have oppositesigns (one positive and the other onenegative).

Fe = −k |q1| |q2|r2 (1)

Physics 202 – p. 14/27



The electrostatic force is repulsive whenq1 q2 > 0, i.e. the two charges have the samesigns (either both positive or both negative).

Fe = k |q1| |q2|r2 (1)

Since Fe goes like 1/r2, the electrostatic forcebecomes stronger for smaller distances andvice versa. Strong bonding at short distancesfor objects such as adhesive tapes.

At the same r, |Fe| � |Fg| (See p. 659).

Physics 202 – p. 15/27



The electrostatic force is repulsive whenq1 q2 > 0, i.e. the two charges have the samesigns (either both positive or both negative).

Fe = k |q1| |q2|r2 (1)

Since Fe goes like 1/r2, the electrostatic forcebecomes stronger for smaller distances andvice versa. Strong bonding at short distancesfor objects such as adhesive tapes.

At the same r, |Fe| � |Fg| (See p. 659).

Physics 202 – p. 15/27


Coulomb’s Force is a vector

~F21 = k q1 q2

r2 r

where r = ~r/r is the unit vector pointing from q1

to q2. ~F21: Force on q2 due to q1.

Physics 202 – p. 16/27


Some examples

Physics 202 – p. 17/27


A problem3 charges lie along the x-axis. Charge q1 = 1µCis at the origin. Charge q2 = −3µC is atx = −0.30m. Charge q3 = −4µC is atx = +0.20m. Find the net electrostatic forceacting on q1.

Concept: Net force on q1= Vector sum of theelectrostatic forces on q1 due to q2 and q3 ⇒Superposition Principle.

Draw a picture of the forces

Physics 202 – p. 18/27


A problem3 charges lie along the x-axis. Charge q1 = 1µCis at the origin. Charge q2 = −3µC is atx = −0.30m. Charge q3 = −4µC is atx = +0.20m. Find the net electrostatic forceacting on q1.

Concept: Net force on q1= Vector sum of theelectrostatic forces on q1 due to q2 and q3 ⇒Superposition Principle.

Draw a picture of the forces

Physics 202 – p. 18/27


A problem: Solution

q1 q2 = −3(µC)2 = −3 × 10−12C2 is negative.The force on q1 from q2 is attractive and pointsin the negative x-direction (toward q2). Let theunit vector pointing in the positive x-directionbe i. One obtains~F12 = −(8.99 × 109 N.m2/C2)3×10−12C2

(0.30m)2 i =

−0.30 Ni.

Physics 202 – p. 19/27


A problem: Solution

Similarly, q1 q3 = −4(µC)2 = −4 × 10−12C2 isnegative. The force on q1 from q3 is attractiveand points in the positive x-direction (towardq3). One obtains~F13 = (8.99×109 N.m2/C2)4×10−12C2

(0.20m)2 i = 0.90 Ni.

Physics 202 – p. 20/27


A problem: Solution

The net force on q1 is~Fnet = ~F12 + ~F13 = 0.60 Niand points in the positive x-direction.

Summary: 1) Calculate each forceseparately; 2) Sum the 2 forces as vectors toget the final result.

Physics 202 – p. 21/27


A problem: Solution

The net force on q1 is~Fnet = ~F12 + ~F13 = 0.60 Niand points in the positive x-direction.

Summary: 1) Calculate each forceseparately; 2) Sum the 2 forces as vectors toget the final result.

Physics 202 – p. 21/27


Coulomb’s Law: Charge distributions

Coulomb’s law also applies to chargedistribution: Same formula.

Example: Charge distributed uniformly overthe volume of a sphere of radius R with avolume charge density ρ ⇒ Total chargeQ = ρ(4

3πR3).

Electrostatic force between 2 such spheres:

~Fe = kQ1 Q2

r2 r

Physics 202 – p. 22/27





3πR3).


~Fe = kQ1 Q2

r2 r

Physics 202 – p. 22/27





3πR3).


~Fe = kQ1 Q2

r2 r

Physics 202 – p. 22/27


Electric field

Recall:~F21 = k q1 q2

r2 r (1)

Define an electric field due to q1:

~E(r) = k q1

r2 r

Rewrite (1) as

~F21 = q2~E(r) (1)

This is the force on q2 in the presence of anelectric field ~E(r)

Physics 202 – p. 23/27


Electric field


r2 r (1)


~E(r) = k q1

r2 r

Rewrite (1) as

~F21 = q2~E(r) (1)


Physics 202 – p. 23/27


Electric field


r2 r (1)


~E(r) = k q1

r2 r

Rewrite (1) as

~F21 = q2~E(r) (1)


Physics 202 – p. 23/27


Electric field


r2 r (1)


~E(r) = k q1

r2 r

Rewrite (1) as

~F21 = q2~E(r) (1)


Physics 202 – p. 23/27


Electric field of point charge: ~E(r) = k qr2 r

Physics 202 – p. 24/27


Electric field: Example problemTwo positive point charges, q1 = +16µC andq2 = +4µC, are separated in a vacuum by adistance of 3.0m. Find the spot on the linebetween the two charges where the net electricfield is zero.

The electric field is a vector.

The net electric field is the vector sum of thetwo individual electric fields.

Physics 202 – p. 25/27


Electric field: Example problemTwo positive point charges, q1 = +16µC andq2 = +4µC, are separated in a vacuum by adistance of 3.0m. Find the spot on the linebetween the two charges where the net electricfield is zero.

The electric field is a vector.

The net electric field is the vector sum of thetwo individual electric fields.

Physics 202 – p. 25/27


Electric field: Example problem

Since both charges are positive, the electricfields coming from q1 and q2 at any point inbetween will be pointing in the oppositedirection to each other. Therefore the netelectric field will be the difference of the twoindividual electric fields.

Physics 202 – p. 26/27


Electric field: Example problem

Let l be the distance from q1 of the point inbetween where the two electric fields exactlycancel each other, resulting in a zero netelectric field. One hask 16µC

l2= k 4µC

(3.0m−l)2

4.0(3.0m − l)2 = l2

2.0(3.0m − l) = ±ll = 2.0m; l = 6.0mOnly l = 2.0m is the correct answer (inbetween point).

Physics 202 – p. 27/27

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Physics 202 - University of Virginiagalileo.phys.virginia.edu/~pqh/202_1n.pdf · Electric Charges, Forces and Fields A problem: 3 charges lie along the x-axis. Charge q1 = 1 C is