Voltage and Capacitance

Chapter 29

Electric Potential Energy

Potential Energy of a charge

• Wants to move when it has high PE

• Point b– U = max– K = min

• Point a– U = min– K = max

Electric Work

Charge moving between plates.

Work = Fr cos0o = qExf - qExi

U = qExf - qExi

A 2.0 cm X 2.0 cm capacitor has a 2.0 mm spacing and a charge of +1.0 nC.

a. Calculate the electric field of the capacitor. (2.82 X 105 N/C)

b. A proton is released from rest at the positive plate. Calculate the change in potential energy. (9.02 X 10-

17 J)

c. Calculate the final speed of the proton. (3.29 X 105 m/s)

d. An electron is released from the halfway point between the plates. Calculate the change in PE and the final speed of the electron. (9.95 X 106 m/s)

A 2.0 cm diameter disk capacitor has a 2.5 mm spacing and a charge of +0.75 nC.

a. Calculate the electric field of the capacitor.

b.An electron is released from rest at the negative plate. Calculate the change in potential energy. (9.02 X 10-17 J)

c. Calculate the final speed of the electron. (3.29 X 105 m/s)

Potential Energy of point charges

Uelec = 1 q1q2

4o r

k = 9.0 X 109 Nm2/C2

A proton is fired from “far away” at a 1.0 cm diameter glass sphere of charge +100 nC.

a. Calculate the initial potential energy of the system (just as it touches the sphere).

b.Since PE = KE, calculate the needed initial speed of the proton.

An electron and a positron are created in the CERN collider.

a. Calculate the potential energy they have when they are 1.0 X 10-10 m apart.

b.Calculate the velocity they need to escape from one another. Remember that PE = KE, but you will need to consider the KE of both particles added together.

Electric Potential: Voltage

• Voltage

• 1 Volt = 1 Joule/Coulomb

V = U

q

Vab = Va – Vb = -Wba

q

Work done by the electric field to accelerate the charge

• The higher the rock, the greater the PE

• The greater the Voltage or charge, the greater the PE (U = qV)

An electron is accelerated in a TV tube through a potential difference of 5000 V.

a. Calculate the change in PE of the electron (-8.0 X 10-16 J)

b.Calculate the final speed of the electron (m = 9.1 X 10-31 kg) (4.2 X 107 m/s (1/7th speed of light)

Calculate the final speed of a proton (mass =

1.67 X 10-27 kg) (9.8 X 105 m/s (0.3% speed of light)

Equipotential Lines

• Equipotential lines are perpendicular to electric field lines

• Voltage is the same along equipotential lines

• Like contour (elevation) lines on a map

Equipotential lines for point charges

Electric Field and Voltage

U = qEs

V = U/q

V = Es

Greater the distance between plates, the greater the voltage

The greater the E field, the greater the voltage

Voltage increases as you move between plates

What is the electric field between two plates separated by 5.0 cm with a voltage of 50V. (1000 V/m)

A capacitor is constructed of 2.0 cm diameter disks separated by a 2.0 mm gap, and charged to 500 V.

a. Calculate the electric field strength (V = Ed)

b.Calculate the charge on each plate (E = Q/0A)

c. A proton is shot through a hole in the negative plate towards the positive plate. It has an initial speed of 2.0 X 105 m/s. Does is have enough energy to reach the other side? (V = PE/q)

Electron Volt

• Energy an electron gains moving through a potential difference of 1 V

1 eV = 1.6 X 10-19 J

• Ex: An electron moving through 1000 V would gain 1000 eV of energy

Voltage due to a Point Charge

• Voltage is not directional (scalar)

• Charged particles (i.e.: electrons, protons) have a voltage

V = kQ

r

Example 1

Consider a +1.0 nC charge.

a. Calculate the electric potential (voltage) at a point 1.0 cm from the charge

b.Calculate the electric potential at a point 3.0 cm from the charge.

Point Charges: Example 2

Calculate voltage (electric potential) at point A as shown below:

A

Q1 = +50 C Q2 = -50 C

30 cm

52 cm

Use Pythagoream theorem to calculate the distance from A to Q2:

A

Q1 = +50 C Q2 = -50 C

30 cm

52 cm

VA = V1 + V2

V1 = kQ = (9.00X 109 Nm2/C2)(5.00X10-5C)

r (0.30 m)

V1 = 1.50 X 106 V

V1 = kQ = (9.00X 109 Nm2/C2)(-5.00X10-5C)

r (0.60 m)

V2 = -7.5 X 105 V

VA = 1.50 X 106 V -7.5 X 105 V

VA = 7.5 X 105 V

Point Charges: Example 3

Calculate voltage (electric potential) at point B as shown below:

B

Q1 = +50 C Q2 = -50 C

30 cm

26 cm 26 cm

Use Pythagoream theorem to calculate the distance from B to Q1 and to Q2:

B

Q1 = +50 C Q2 = -50 C

30 cm

26 cm 26 cm

VA = V1 + V2

V1 = kQ = (9.00X 109 Nm2/C2)(5.00X10-5C)

r (0.40 m)

V1 = 1.125 X 106 V

V1 = kQ = (9.00X 109 Nm2/C2)(-5.00X10-5C)

r (0.40 m)

V2 = -1.125 X 105 V

VA = 1.125 X 106 V –1.125 X 105 V

VA = 0 V

Point Charges: Example 4

How much work is required to bring a charge of q = 3.00 C to a point 0.500 m from a charge Q = 20.0 C?

VQ = kQ = (9.00X 109 Nm2/C2)(2.00X10-5C)

r (0.500 m)

VQ = 3.6 X 105 V (This is the voltage caused by the stationary charge)

W = U (like the work to lift a book to a shelf)

V = U

q

V = W

q

W = Vq

W = (3.6 X 105 V )(3 X 10-6 C)

W = 1.08 J

Point Charges: Example 5

Which of three sets of charges has the most:

• positive potential energy?

• The most negative potential energy?

• Would require the most work to separate?

+

+

+

-

-

+

(i)

(ii)

(iii)

Charged Spheres

Act as point charges

At surface of sphere

Substitute:

A proton is released from rest at the surface of a 1.00 cm-diameter sphere charged to +1000 V.

a. Calculate the charge on the sphere (0.56 nC)

b.Calculate the speed of the proton when it is 1.00 cm from the sphere (note that it has potential energy in both cases, U = qV) (3.6 X105 m/s)

A thin ring has a radius R and charge Q. Find the potential at a distance of z from the axis of the ring.

A thin disk has a radius R and charge Q. Find the potential at a distance of z from the axis of the ring.

A 17.5 mm diameter dime is charged to +5.00 nC.

a. Calculate the potential of the dime (10,300 V)

b.Calculate the potential 1.00 cm above the dime (3870 V)