Question

The potential difference between two points separated by 50 cm in the field of source charge 500 ×10^-6 is 400 V. The electric field intensity of the charge is
a. 20000 N/V
b. 400 V/m
c. 2000 V/m
(correct answer is (a). kindly give proper reason with solution otherwise don't answer)​

Answer 1

answer is a 20000

Explanation:

first calculate potential energy from formula U=V/q

here U comes out to be 800000

and then using relation b/w U and E

we can find out feild intensity