Potential Difference and Capacitance

Consider a rock dropped in a grav. field:

Work is done by gravity

To lift the object, Work is done to give the rock PE!

High PE

Low PE

A test charge in an elect. Field:+ + + + + + + + + + + + +

– – – – – – – – – – – –

q

q

Work is done by the electric field -- a difference in PE exists

Potential Difference [electric potential] (V): Work done in moving a charge across an electric field per unit of charge.

≈ Similar (but not exactly) to PE is Work done in moving an object up through a gravitational field

V = W

q

Units:volt (v) = J/C

Different masses, same h, different PE!

+ + + + + + + + + + + + +

– – – – – – – – – – – –

q 2q

Not same PE-- the 2q charge has greater PE

However, same Potential Difference:

V = W = 2W

q 2q

To move a larger charge across an electric field will require more work– but the potential

difference of the field (voltage) stays constant!

Some Typical Voltages (approximately):

Thundercloud to ground… 108 V

High Power Line…. 106 V

Television Power Supply…. 104 V

Household outlet…. 1.1 X 102 V (in America)

Potential changes on skin (EEG and EKG)…. 10-4 V

1) How much work is needed to move a charge of -8.6 µC to a point where the potential difference is +75.0 V?

2) How much work is needed to move a proton from a point of potential of +100.0 V to a potential of -50.0 V?3) An electron falls through a potential difference of 21,000 V in a TV tube. How much kinetic energy will it gain?

4) A force of 6.4 N is required to move a charge of 12.3 µC across an electric field created by two charged plates 3.00mm apart. What is the potential difference across these plates?

Capacitance

A capacitor is a very common device used in electronics to store electric charge.

• camera flashes, RAM, surges, etc.

• usually two parallel plates (often rolled)

+ +

+ +

+ +

+ – – – – – – –

Capacitance (C) : the ratio of the charge on either plate to the potential difference between the plates

C =Q

V

Units:

F (farad) =C

V

Capacitors have much less capacitance than a single farad:

1 µF = 10-6 F and 1 pF = 10-12

The insulating material between the plates will also affect the capacitance:

+ + + + +

- - - - -

Air

+ + + + + + + + + + +

- - - - - - - - - - -

glass

C1 C2

V1 = V2 but Q2 = KQ1 Therefore:

C2 = KC1 where K is called the dialectric constant

Dialectric constants are ratios compared to a vacuum!

Material Dielectric Constant

Paper 1.5 to 2.6

Paraffin 2.0 to 2.5

Rubber 2.0 to 3.5

Glass 5.4 to 10

Water 80

Special electrical paper and plastics are commonly used dielectric materials. Their dielectric constants will vary depending on how they are manufactured.

Combinations of Capacitors

Capacitors can be connected in parallel or in series:

Parallel: All positive plates are connected to each other and all negative plates are connected to each other

Series: Positive plates connected to negative plates

+

-

+

+

+

-

-

-

For capacitors connected in parallel, the larger the capacitance, the larger the stored charge, so the potential difference is constant:

Q1 = C1V, Q2 = C2V, Q3 = C3V

QT = CTV

QT = Q1 + Q2 + Q3

CTV = C1V + C2V + C3V

V cancels out, so for capacitors in Parallel:

CT = C1 + C2 + C3

For capacitors connected in series, the charge on each is constant, so:

QT = Q1 = Q2 = Q3

The potential difference will vary with capacitance and the total across the combination will be:

VT =QT

CT

VT = V1 + V2 + V3

Q = Q + Q + Q

CT C1 C2 C31 = 1 + 1 + 1

CT C1 C2 C3

Four capacitors are connected in parallel and charged to a potential difference of 150.0 V. The capacitor values are .300 µF, .300 µF, .400 µF, and .400 µF. What is the charge on one of the .400 µF capacitors?

VT = 150.0 VC1 = .300 µFC2 = .300 µFC3 = .400 µFC4 = .400 µF

Q3 = ?

Q3 = C3V3

VT = V1 = V2 = V3 …

Q3 = (.400 µF)(150.0 V)

= 60.0 µC

The same four capacitors are discharged and connected in series. A charge of 60.0 µC is measured across one of the .400 µF capacitors. Calculate the potential difference across one of the .300 µF capacitors.

C1 = .300 µFC2 = .300 µFC3 = .400 µFC4 = .400 µF

Q3 = 60.0 µCV2 = ?

V2 =Q2

C2

Q = Q1 = Q2 = Q3….

V2 = 60.0 µC/.300 µF = 200 V

1) What is the potential difference between two points if .0065 J is required to move a charge of 3.2 µC between the points?2) How much work must a 1.5 V battery do if it is to move .25 C per second through a wire for 13 minutes?3) What is the equivalent capacitance when a 2.0 µF, a 4.0 µF and a 6.0 µF are connected in A) series B) parallel?4) A proton is accelerated through a potential difference of 4.5 x 106 V. A) How much kinetic energyhas it acquired? B) If it started from rest, how fast is it going?

5) Given three 2.0 µF capacitors, draw all possible circuit diagrams and calculate the equivalent capacitance of each combination.

6) A pair of parallel plates is .0500 cm apart. Their capacitance with air between them is 75.0 pF. A) What is their capacitance with air replaced with paraffin? [ K = 2.2 ] B) If the potential difference between the plates is 25.5 V, what is the charge on each plate with paraffin? C) What is the equivalent capacitance of two pairs of these plates connected in series? D) What is the equivalent capacitance of two pairs of these plates connected in parallel?