Question

Two charges 2uC& 8uC are separated by 2mm in air. Find the points of zero electric
intensities on the line joining the two charges the two charges​

Answer 1

Answer :

23m from 4μC

Solution :

Here, q1=4μC,q2=−2μC

Let the potential be zero at a point P, at a distance r1=x from 4μC charge, Therefore, r2=(I−x)

Potential at P=kq1r1+kq2r2=0

∴q1r1=−q2r2

4x=−(−2)1−x

x=2−2x

3x=2,x=23m from 4μC charge