There are two equipotential surface as shown in figure. The distance between them is $r$. The charge of $-q\,$ coulomb is taken from the surface $A$ to $B$, the resultant work done will be
$W = \frac{1}{{4\pi {\varepsilon _o}}}\frac{q}{r}$
$W = \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{{{r^2}}}$
$W = - \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{{{r^2}}}$
$W = zero$
A test charge $q$ is made to move in the electric field of a point charge $Q$ along two different closed paths as per figure. First path has sections along and perpendicular to lines of electric field. Second path is a rectangular loop of the same area as the first loop. How does the work done compare in the two cases ?
If identical charges $( - q)$ are placed at each corner of a cube of side $b$, then electric potential energy of charge $( + q)$ which is placed at centre of the cube will be
A point charge is surrounded symmetrically by six identical charges at distance $r$ as shown in the figure. How much work is done by the forces of electrostatic repulsion when the point charge $q$ at the centre is removed at infinity
A particle $A$ has charge $+q$ and particle $B$ has charge $+4 q$ with each of them having the same mass $m$. When allowed to fall from rest through the same electric potential difference, the ratio of their speeds $\frac{V_A}{V_B}$ will become
The diagram shows three infinitely long uniform line charges placed on the $X, Y $ and $Z$ axis. The work done in moving a unit positive charge from $(1, 1, 1) $ to $(0, 1, 1) $ is equal to