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
$W = \frac{1}{{4\pi {\varepsilon _o}}}\frac{q}{r}$
$W = \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{{{r^2}}}$
$W = - \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{{{r^2}}}$
$W = zero$
Two points $P$ and $Q$ are maintained at the potentials of $10\, V$ and $-4\,V$, respectively. The work done in moving $100$ electrons from $P$ and $Q$ is
When one electron is taken towards the other electron, then the electric potential energy of the system
A point chargr $Q$ is fixed A small charge $q$ and mass $m$ is given a velocity $v_0$ from infinity & perpendicular distance $r_0$ as shown. If distance of closest approach is $r_0/2$. The value of $q$ is [Given $mv_0^2 = \frac{{{Q^2}}}{{4\pi { \in _0}\,{r_0}}}$]
A proton of mass $m$ and charge $e$ is projected from a very large distance towards an $\alpha$-particle with velocity $v$. Initially $\alpha$-particle is at rest, but it is free to move. If gravity is neglected, then the minimum separation along the straight line of their motion will be
Which of the following statement$(s)$ is/are correct?
$(A)$ If the electric field due to a point charge varies as $r^{-25}$ instead of $r^{-2}$, then the Gauss law will still be valid.
$(B)$ The Gauss law can be used to calculate the field distribution around an electric dipole.
$(C)$ If the electric field between two point charges is zero somewhere, then the sign of the two charges is the same.
$(D)$ The work done by the external force in moving a unit positive charge from point $A$ at potential $V_A$ to point $B$ at potential $V_B$ is $\left(V_B-V_A\right)$.