Question

Open in App

Verified by Toppr

Correct option is D)

The potential energy of the system is given by $U=U_{12}+U_{13}+U_{23}$

Here, $1,2,3$ denote the three charges in order respectively.

Let $4πϵ_{0}1 =k$

Then, $U_{12}=kd−qQ $, $U_{13}=k2dq_{2} $ and $U_{23}=kd−qQ $

Thus, the potential energy of the system is $U=2dk (−2qQ+q_{2}−2qQ)=2dk (q_{2}−4qQ)$

But, given that $U=0⇒q_{2}=4qQ⇒Q:q=1:4$

Solve any question of Electrostatic Potential and Capacitance with:-

0

0