Side length of equilateral triangle is $d. P$ is mid of side then potential at point $P, V_P$ is
$\frac{Q}{{2\pi { \in _0}d}}\left[ {2 + \sqrt 3 } \right]$
$\frac{Q}{{2\pi { \in _0}d}}\left[ {2 + \frac{1}{{\sqrt 3 }}} \right]$
$\frac{Q}{{4\pi { \in _0}d}}\left[ {4 + \sqrt 3 } \right]$
Zero
Consider the situation shown. The switch $S$ is opened for a long time and then closed. The charge flown through the battery when $S$ is closed
Two equal point charges are fixed at $x = -a$ and $x = + \,a$ on the $x$-axis. Another point charge $Q$ is placed at the origin. The change in the electrical potential energy of $Q$ ehen it is displaced by a small distance $x$ along the $x$ -axis is apporximately proportional to
Three capacitors $1, 2$ and $4\,\mu F$ are connected in series to a $10\, volts$ source. The charge on the plates of middle capacitor is
Four point $+ve$ charges of same magnitude $(Q)$ are placed at four corners of a rigid square frame in $xy$ plane as shown in figure. The plane of the frame is perpendicular to $z-$ axis. If a $-ve$ point charges is placed at a distance $z$ away from the above frame $(z << L)$ then
A $2\,\mu F$ capacitor is charged as shown in the figure. The percentage of its stored energy dissipated after the switch $S$ is turned to position $2$, is.....$\%$