Electric Potential Due to a System of Charges

A charge $+Q$ is placed on a large spherical conducting shell of radius $R$. Another small conducting sphere of radius r carrying charge $q$ is introduced inside the large shell and is placed at its centre. The potential difference between two points, one lying on the sphere and the other on the shell would be:

A solid conducting sphere, having a charge $Q,$ is surrounded by an uncharged conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be $V.$ If the shell is now given a charge of $–4Q,$ the new potential difference between the same two surfaces is:

Two thin wire rings each having a radius $R$ are placed at a distance d apart with their axes coinciding. The charges on the two rings are $\mathit{+}Q\mathit{ }$and $\mathit{-}Q\mathit{.}$ The potential difference between the centres of the two rings is :-

A charge $+q$ is distributed over a thin ring of radius $r$ with line charge density $\lambda =\frac{q{\sin }^2\theta }{\pi r}$. Note that the ring is in the $XY$-plane and $\theta$ is the angle made by $r$ with the $X$-axis. The work done by the electric force in displacing a point charge $+Q$ from the centre of the ring to infinity is

A conducting sphere of radius $a$ is enclosed by a conducting shell whose radius of inner surface is $2\mathrm{a}$ and that of outer surface is $3\mathrm{a}$. Charges given to conducting shell and conducting sphere are $\mathrm{Q}$ and $-\mathrm{Q}$ respectively, then what will be potential at centre of conducting sphere :-

$\mathrm{A}$ conducting spherical bubble of radius $r$ and thickness $t \left(t<<r\right)$ is charged to $\mathrm{a}$ potential $V$. If it collapses to form $\mathrm{a}$ small spherical drop, then the potential of the drop is

Two identical rings each of radius $R$ are coaxially placed $\mathrm{a}$ distance $R$ apart. They carry charges $Q_1$ and $Q_2$ respectively. If a charge $q$ is moved from the centre of one ring to the centre of the other ring, the work done is

Two identical thin rings, each of radius $R$ meters, are coaxially placed at a distance $R$ meters apart. If $Q$ coulomb of charges is uniformly spread over both the rings, the work done in moving a charge $q$ from the centre of one ring to that of other is:

A spherical conductor of radius $3 \mathrm{m}$ is charged to a potential of $90 \mathrm{V}$. It is now placed inside another hollow spherical conductor of radius $6 \mathrm{m}.$Calculate the potential of bigger sphere, if the smaller sphere is made to touch the bigger sphere.

A charge Q is distributed over two concentric hollow spheres of radius r and R (< r) such that the surface densities are equal. The potential at the common centre is :

Figure shows a annular disc of outer radius $R$ inner radius $R,$ and uniform surface charge density. Total charge is $q.$ The potential at the centre is $\frac{kq}{r^{\text{'}}}\left(k=\frac{1}{4 \pi {\epsilon }_0}\right)$ where $r^{\text{'}}$ is

Three charges $-q,+q$ and $+q$ are situated in $X-Y$ plane at points $(0,-\mathrm{a}), (0,0)$ and $(0,\mathrm{a})$ respectively. The potential in $xy$ plane at a distant point $r\left(r>>a\right)$ in a direction making an angle $\theta$ from the Y-axis:-

An arc of radius $R$ is shown in the figure, find the potential at its centre:-

The radius of hollow metallic sphere is $r$. If the potential difference between its surface and a point at a distance $3r$ from its centre is $V,$ then the electric field intensity at a distance of $3r$ from its centre is:-

As one moves towards the centre from surface of uniformly charged metallic sphere the potential-

$1000$ small water drops each of radius $r$ and charge $q$ coalesce together to form one spherical drop. The potential of the big drop is larger than that of the smaller drop by a factor of:-

A solid conducting sphere having a charge $Q$ is surrounded by an uncharged concentric conducting hollow shperical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be $\mathrm{V}.$ If the shell is now given a charge of $–3Q,$ the new potential difference between the same two surface is:-

Half of the non-conducting ring has $\left(+Q\right)$ charge and half has $\left(-Q\right)$ charge. Find potential at point $A$ which is on the line passing through centre perpendicular to plane of ring.

A charge $Q$ is uniformly distributed through out the volume of a dielectric sphere of radius $R$. If potential at its surface is $24$ volts. Then potential at $r=\frac{R}{2}$ and $r=\frac{3R}{2}$ respectively are.... ...

(where $r$ distance from the centre of sphere):-