Question

The potential difference between two points separated by 50cm in the field of source charge 500 microC is 400V. The electric field intensity of the charge is?
A- 20000 N/V
B- 400V/m
C- 800V/m
D- 2000 N/V

Answer 1

The potential difference between two points

$$V=Ed$$

Hence, the electric field intensity of the charge

$$E=\frac{V}{d}=\frac{400\:\rm V}{0.50\:\rm m}=800\:\rm V/m$$

Answer: C