question_answer

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]

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 m