Equal charges are given to two spheres of different radii. The potential will
Be more on the smaller sphere
Be more on the bigger sphere
Be equal on both the spheres
Depend on the nature of the materials of the spheres
As shown in the figure, charges $ + q$ and $ - q$ are placed at the vertices $B$ and $C$ of an isosceles triangle. The potential at the vertex $A$ is
The linear charge density on a dielectric ring of radius $R$ varies with $\theta $ as $\lambda \, = \,{\lambda _0}\,\cos \,\,\theta /2,$ where $\lambda _0$ is constant. Find the potential at the centre $O$ of ring. [in volt]
A charge $ + q$ is fixed at each of the points $x = {x_0},\,x = 3{x_0},\,x = 5{x_0}$..... $\infty$, on the $x - $axis and a charge $ - q$ is fixed at each of the points $x = 2{x_0},\,x = 4{x_0},x = 6{x_0}$,..... $\infty$. Here ${x_0}$ is a positive constant. Take the electric potential at a point due to a charge $Q$ at a distance $r$ from it to be $Q/(4\pi {\varepsilon _0}r)$. Then, the potential at the origin due to the above system of charges is
A charge of total amount $Q$ is distributed over two concentric hollow spheres of radii $r$ and $R ( R > r)$ such that the surface charge densities on the two spheres are equal. The electric potential at the common centre is
A solid sphere of radius $R$ is charged uniformly. At what distance from its surface is the electrostatic potential half of the potential at the centre?