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 charge of $8\; mC$ is located at the origin. Calculate the work done in $J$ in taking a small charge of $-2 \times 10^{-9} \;C$ from a point $P (0,0,3\; cm )$ to a point $Q (0,4\; cm , 0),$ via a point $R (0,6\; cm , g \;cm )$
In the figure, the inner (shaded) region $A$ represents a sphere of radius $r_A=1$, within which the electrostatic charge density varies with the radial distance $r$ from the center as $\rho_A=k r$, where $k$ is positive. In the spherical shell $B$ of outer radius $r_B$, the electrostatic charge density varies as $\rho_{\bar{B}}=\frac{2 k}{r}$. Assume that dimensions are taken care of. All physical quantities are in their $SI$ units.
Which of the following statement($s$) is(are) correct?
Three charges are placed along $x$-axis at $x=-a, x=0$ and $x=a$ as shown in the figure. The potential energy of the system is
A ball of mass $1\, g$ and charge ${10^{ - 8}}\,C$ moves from a point $A$. where potential is $600\, volt$ to the point $B$ where potential is zero. Velocity of the ball at the point $B$ is $20\, cm/s$. The velocity of the ball at the point $A$ will be
A particle of mass $100\, gm$ and charge $2\, \mu C$ is released from a distance of $50\, cm$ from a fixed charge of $5\, \mu C$. Find the speed of the particle when its distance from the fixed charge becomes $3\, m$. Neglect any other force........$m/s$