5. The electric field between the plates of an ideal parallel plate capacitor ha

Question

5. The electric field between the plates of an ideal parallel plate capacitor has magnitude 2000 V/m. What is the change in potential energy of the system as a -1C charge is moved from A to B? (See Graphic Above)
a) +60 J
b) +20 J
c) -20 J
d) -60 J
e) -100 J
6. When a -2 C charge is moved from point A to B, the potential energy of the system increases by 10 J. What is the voltage VAB?
a) -10 V
b) -5 V
c) 0 V
d) +5 V
e) +10 V

7. The potential at the surface of a solid conducting sphere of radius 0.4 m is 12 Volts. All charges on
the sphere remain at rest. What is the potential at a point midway between surface and center.?
a) 0 V
b) 4 V
c) 8 V
d) 12 V
e) 16 V

IF hand written please write neat and clear because it is hard to read sometimes. Please show formulas used. Thank you

Explanation / Answer

5)   E = 2000 V/m

      q = -1 C

      d = 0.03 m

Change in potential energy = U = qEd = -1 C*2000 V/m*0.03 m = -60 J (Option D)

6) q = -2C

      U = 10 J

Now   U = qV =====> V = U/q = 10/-2 = -5 V   (Option B)

7) Potential inside the sphere will remain same = 12 V (Option D)

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