In moving from $A$ to $B$ along an electric field line, the electric field does $6.4 \times {10^{ - 19}}\,J$ of work on an electron. If ${\phi _1},\;{\phi _2}$ are equipotential surfaces, then the potential difference $({V_C} - {V_A})$ is.....$V$
$-4$
$4$
$0$
$64$
A particle of mass $‘m’$ and charge $‘q’$ is accelerated through a potential difference of $V$ volt, its energy will be
As shown in figure, on bringing a charge $Q$ from point $A$ to $B$ and from $B$ to $C$, the work done are $2\, joule$ and $-3\, joule$ respectively. The work done to bring the charge from $C$ to $A$ is
A solid sphere of radius $R$ carries a charge $(Q+q)$ distributed uniformly over its volume. A very small point like piece of it of mass $m$ gets detached from the bottom of the sphere and falls down vertically under gravity. This piece carries charge $q.$ If it acquires a speed $v$ when it has fallen through a vertical height $y$ (see figure), then :
(assume the remaining portion to be spherical).
Derive the formula for the electric potential energy of system of two charges.
$(a)$ Calculate the potential at a point $P$ due to a charge of $4 \times 10^{-7}\; C$ located $9 \;cm$ away.
$(b)$ Hence obtain the work done in bringing a charge of $2 \times 10^{-9} \;C$ from infinity to the point $P$. Does the answer depend on the path along which the charge is brought?