question_answer

Two spheres carrying charges \[+6\mu C\] and \[+9\mu C\], separated by a distance d, experiences a force of repulsion F. When a charge of \[-3\mu C\]is given to both the sphere and kept at the same distance as before, the new force of repulsion is

A)  \[3F\]    

B)  \[\frac{F}{9}\] 

C)  \[F\]     

D)  \[\frac{F}{3}\]

Correct Answer: D

Solution :

: Let us call spheres as A and B. According to Coulomb s law, the force of repulsion between A and B separated by a distance d is \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(+6\mu C)(+9\mu C)}{{{d}^{2}}}\]       ??(i) When a charge of \[-3\mu C\] is given to both the spheres, then charge on \[A=+6\mu C-3\mu C\] \[=+3\mu C\] and on s\[B=+9\mu C-3\mu C=+6\mu C\] Again by Coulombs law, the new force of repulsion between A and B separated by the same distance d is \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(+3\mu C)(+6\mu C)}{{{d}^{2}}}\] Dividing eqn. (ii) by eqn. (i), we get \[\frac{F}{F}=\frac{\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(+6\mu C)(+9\mu C)}{{{d}^{2}}}}{\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(+6\mu C)(+9\mu C)}{{{d}^{2}}}}\] \[=\frac{(+3\mu C)(+6\mu C)}{(+6\mu C)(+9\mu C)}=\frac{1}{3}\] or \[F=\frac{F}{3}\]