First slide

Electrostatic potential

Question

A charge of 5 C experiences a force of 5000 N when it is kept in a uniform electric field. What is the potential difference between two points separated by a distance of 1 cm?

Easy

A

10 V
B

250 V
C

1000 V
D

2500 V

Solution

$F = QE = \frac{QV}{d} \Rightarrow 5000 = \frac{5 \times V}{10^{- 2}} \Rightarrow V = 10 volt$

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