Question

The surfaces of two large parallel conducting plates separated by 5.0 cm have uniform surface charge densities that are equal in magnitude but opposite in sign. The difference in potential between the plates is 200 V. (a) Is the positive or the negative plate at the higher potential? (b) What is the magnitude of the electric field between the plates? (c) An electron is released from rest next to the negatively charged surface. Find the work done by the electric field on the electron as the electron moves from the release point to the positive plate. (d) What is the change in potential energy of the electron when it moves from the release point plate to the positive plate? Is the potential energy increasing or decreasing?

Answer 1

Part A.

Positive plate willbe at higher potential because electric field will be directed from +ve to -ve plate which means electric potential will be higher at positive plate, (Since E = -dV/dx)

Part B.

Relation between electric field and potential difference between plates is given by:

E = V/d

V = Potential difference = 200 V

d = plate separation = 5.0 cm = 0.05 m

So,

E = 200/0.05 = 4000 V/m

Part C.

Work-done in moving a charge through potential difference is given by:

W = q*dV

q = charge on electron = 1.6*10^-19 C

So,

W = 1.6*10^-19*200

W = 3.2*10^-17 J = (3.2*10^-17/(1.6*10^-19)) eV = 200 eV

W = 3.2*10^-17 J = 200 eV

Part D.

Relation between work-done and change in potential energy is given by:

W = -dU, So

dU = -200 eV = -3.2*10^-17 J

dU = Uf - Ui, Since dU is -ve, So potential energy is decreasing.

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