Q. Two point charges $+ 8q$ and $- 2q$ are located at $x = 0$ and $x = L$ respectively. The location of a point on the $x$-axis at which the net electric field due to these two point charges is zero is
A) $8\ L$
B) $4\ L$
C) $2\ L$
D) $\frac{L}{4}$
Explanation
(c) The net field will be zero at a point outside the charges and near the charge which is smaller in magnitude
Suppose E.F. is zero at $P$ as shown
Hence at $P;k.\frac{8q}{(L + l)^{2}} = \frac{k.(2q)}{l^{2}} \Rightarrow l = L$
So distance of $P$ from origin is $L + L = 2L$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.