Electrostatics Question 420
Question: Two electric charges $ 12\mu C $ and $ -6\mu C $ are placed 20 cm apart in air. There will be a point P on the line joining these charges and outside the region between them, at which the electric potential is zero. The distance of P from $ -6\mu C $ charge is [EAMCET 2000]
Options:
A) 0.10 m
B) 0.15 m
C) 0.20 m
D) 0.25 m
Show Answer
Answer:
Correct Answer: C
Solution:
Point P will lie near the charge which is smaller in magnitude i.e. -6 m C.
Hence potential at P
$ V=\frac{1}{4\pi {\varepsilon _{0}}}\frac{(-6\times {{10}^{-6}})}{x}+\frac{1}{4\pi {\varepsilon _{0}}}\frac{(12\times {{10}^{-6}})}{(0.2+x)}=0 $
therefore x = 0.2 m
