A) 0.10 m
B) 0.15 m
C) 0.20 m
D) 0.25 m
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 \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(-6\times {{10}^{-6}})}{x}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(12\times {{10}^{-6}})}{(0.2+x)}=0\] Þ x = 0.2 mYou need to login to perform this action.
You will be redirected in
3 sec