Question

The potential due to a point charge at a distance of
0.3 m is 10^3 V. What will be the electric field intensity
due to the same charge at a distance of 0.4 m?
Ans. 1875 V-m​

Answer 1

Answer:

Explanation:

kq/r1=VI

Here k is constant

quality is point charge

r1 is distance from point charge

VI is potential

This implies that

kq=V1r1

Electric field at distance r2. from the point charge quality is

E=kq/(r2)^2=V1r1/(r2)^2

=(10)^3×0.3/(0.4)^2

=(300/16)=75/4

=18.75 V/m