Question1. The electric field strength between two parallel conducting plates se

Question

Question1.

The electric field strength between two parallel conducting plates separated by 5 cm is 7.4 × 104 V/m.

Part (a) What is the potential difference between the plates in kV? Numeric : A numeric value is expected and not an expression.

VAB = __________________________________________

Part (b) The plate with the lowest potential is taken to be at zero volts. What is the potential 1.00 cm from that plate in V? Numeric : A numeric value is expected and not an expression.

VAB = __________________________________________

Explanation / Answer

d = distance between plates = 5 cm = 0.05 m

E = electric field = 7.4 x 104 V/m

a)

Potential difference between plates is given as

V = E d

V = (7.4 x 104) (0.05)

V = 3700 volts

b)

Potential at 1 cm

VAB = E (0.01)

VAB = (7.4 x 104) (0.01)

VAB = 740 volts

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