Answer to Question #128699 in Field Theory for Asif Sultan

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