The charge given to any conductor resides on its outer surface, because
The free charge tends to be in its minimum potential energy state
The free charge tends to be in its minimum kinetic energy state
The free charge tends to be in its maximum potential energy state
The free charge tends to be in its maximum kinetic energy state
Electrostatic potential energy of given system will be
Two charges ${q_1}$ and ${q_2}$ are placed $30\,\,cm$ apart, shown in the figure. A third charge ${q_3}$ is moved along the arc of a circle of radius $40\,cm$ from $C$ to $D$. The change in the potential energy of the system is $\frac{{{q_3}}}{{4\pi {\varepsilon _0}}}k$, where $k$ is
A charged particle $q$ is shot towards another charged particle $Q$ which is fixed, with a speed $v$. It approaches $Q$ upto a closest distance $r$ and then returns. If $q$ were given a speed $2v$, the closest distances of approach would be
If a charge is shifted from a low potential region to high potential region, the electric potential energy