$A$ and $B$ are two identical blocks made of a conducting material. These are placed on a horizontal frictionless table and connected by a light conducting spring of force constant $‘K’$. Unstretched length of the spring is $L_0$. Charge $Q/2$ is given to each block. Consequently, the spring stretches to an equilibrium length $L$. Value of $Q$ is
$\sqrt {4\pi {\varepsilon _0}KL} $
$L\sqrt {\frac{K}{{4\pi {\varepsilon _0}\left( {L - {L_0}} \right)}}} $
$2L\sqrt {4\pi {\varepsilon _0}K\left( {L - {L_0}} \right)} $
$4\pi {\varepsilon _0}K\left( {L - {L_0}} \right)$
Two charges $q$ and $-3q$ are placed fixed on $x-axis$ separated by distance $'d'$. Where should a third charge $2q$ be placed such that it will not experience any force ?
Two identical charged spheres suspended from a common point by two massless strings of lengths $l,$ are initially at a distance $d\;(d < < l)$ apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity $v.$ Then $v$ varies as a function of the distance $x$ between the spheres, as
The unit of electric permittivity is
Write some important points for vector form of Coulomb’s law.
Two identical conducting spheres $A$ and $B,$ carry equal charge. They are separated by a distance much larger than their diameter, and the force between them is $F$ . A third identical conducting sphere, $C,$ is uncharged. Sphere $C$ is first touched to $A,$ then to $B,$ and then removed. As a result, the force between $A$ and $B$ would be equal to