PHY 2054C - H. K. Ng

Chapter 17 - Electric Potential and Electric Energy.

Summary

Work done against electric potential W = q(Vf - Vi), V has unit of volt (V) and W has unit of Joule (J)

For a uniform electric field E = V/d

For point charge, electric potential V = kQ/r, assume V = 0 at r = ¥.

Capacitor is a device to store charges

Q = CV, C is called capacitance (note: it is not the unit for charge, C)

For 2-parallel plates C = KeoA/d, K is called dielectric constant

Energy stored in a capacitor U = ½ QV = ½ CV2 = ½ Q2/C.

________________________________________________________________________

Homework solutions

4. An electron gains KE = 3.45´10-16 J going from plate A to plate B. Find potential difference between plates.

 This is a conservation of energy problem. The gain in KE = loss in potential energy.i.e., KE = qDV = q(Vf - Vi) where q = e = -1.6´10-19 C = -2156 V (volts)

The minus sign tells us that for negative charges, the charges go from low to high potential. Plate B is at a higher potential. (This is opposite that of a positive charge which moves from high to low potential.

6. Require an electric field of E = 640 V/m. Distance between two parallel plates d = 11.0 m. Find voltage between the plates.

For 2 parallel plates where the electric field is uniform

V = Ed

= (640 V/m)(11.0´10-3 m)

= 7 V

9. Work done by an external force Wext = 25´10-4 J. Charge q = -7.5 mC, initially at rest and has 4.82´10-4 J of KE at b, find potential difference.

 There is no given formula you can pluck into to solve this problem. The work done by the external force goes into overcoming the electric potential plus making the charge move, i.e., into kinetic energy.

Wext = Wba + KE

Wba = Wext - KE

= 25.0´10-4 J – 4.82´10-4 J

= 20.18´10-4 J

But Wba = qVba

Therefore Vba = (20.18´10-4 J)/(-7.5´10-6 C)

= -269 V

Thus potential between a and b, Vab = -Vba = +269 V. (Point a has higher potential.)

15.

 The work done is the charge moved times the change in electric potential between the 2 points where the charge was moved. The first step is to find the electric potential at the mid-point and 10 cm away due to the two charges at the ends.

For a point charge, the potential is

Potential at mid-point

= 3.375´106 V

The potential 10 cm away from mid-point is

= 5.54´106 V

Change in potential Vba = Vb - Va = 5.54´106 V – 3.38´106 V = 2.16´105 V

Work done against potential is Wba = qVba

= (0.5´10-6 C)(2.16´106 V)

= 1.08 J.

16. a. The proton is treated as a point charge with Q = 1.6´10-19 C. For a point charge

= 5.76´105 V

1. The electric potential energy of 2 protons that is 2.5´10-15 m apart.

The energy is obtained by moving one proton from infinity to a distance of 2.5´1015 m away. At infinity the electric potential due to the first proton is zero.

Therefore work Wba = qVba = e(Vb - V@¥)

= +1.6´10-19 C(5.76´105 V - 0)

= 9.23´10-14 J

17. To penetrate the carbon atom, the proton must first overcome the electric potential of the C atom. Just as the proton touches the carbon atom, the distance between them is

 Total distance between centers is 1.2´10-15 m + 3.6´10-15 m = 4.8´10-15 m.

Thus potential

= 4.2´106 V = 4.2 MV.

35. Pair of circular plates r = 5.0 cm, distance between plates d = 3.2 mm filled with mica.

The capacitance for //-plate capacitor is given by

For mica K = 7

= 1.5´10-10 F

= 0.15 nF.

38. Capacitor charge Q = 4.2 mC

Electric field E = 2.0 kV/mm = 2.0´106 V/m

Separated by air d = 4.0 mm, find area of the plate.

The area comes from C = eoA/d; use eo instead of e since air in between plates.

We know eo and d but not C or A.

But Q = CV = CEd

\ C =Q/Ed

Hence C =Q/Ed = eoA/d

or

40. Given dielectric constant K = 3.75, electric field E = 9.21´104 V/m, distance between plates d = 1.95 mm, and charge Q = 0.775 mC. Find capacitance C and area of each plate.

One of the expressions for capacitance is

C = Q/V where V = Ed = (9.21´104 V/m)(1.95´10-3 m) = 180 V.

Thus, capacitance

C = (0.775´10-6 C)/(180 V) = 4.31´10-9 F = 4.31 nF.

To find the area of each plate use

C = K eoA/d.

Thus A = Cd/Keo = (4.31´10-9 F)(1.95´10-3 m)/(3.75)(8.85´10-12 C2/N×m2)

= 0.24 m2.

45. Capacitor diameter D = 9.0 inches = 22.9 cm, \ radius r = 0.1145 m

Distance between plates d = 10 cm

Voltage V = 9.0 V

1. Estimate capacitance

= 3.6´10-12 F

= 3.6 pF.

1. Charge on each plate

Q = CV = (3.6´10-12 F)(9V)

= 3.3´10-11 C (Unit for charge Q is Coulomb)

1. Electric field halfway between the plate

For parallel plate capacitor the electric field is uniform between the plates

i.e., E = V/d = (9 V)/(0.1 m) = 90 V/m

halfway (or ¼ or ¾ or …) between the plates.

1. Work done by battery to charge up the plates

W = ½ qV = ½ (3.3´10-11 C)(9 V) = 1.5´10-10 J.

1. Insert a dielectric

C changed since C = eA/d

Q changed

E unchanged but W increase

Thus C, Q and W increase but E is unchanged.

Access counter: times since January 1, 1997.