Consider a sphere of radius $R$ with uniform charge density and total charge $Q$. The electrostatic potential distribution inside the sphere is given by $\theta_{(r)}=\frac{Q}{4 \pi \varepsilon_{0} R}\left(a+b(r / R)^{C}\right)$. Note that the zero of potential is at infinity. The values of $(a, b, c)$ are

Two charges of $4\,\mu C$ each are placed at the corners $A$ and $B $ of an equilateral triangle of side length $0.2\, m $ in air. The electric potential at $C$ is $\left[ {\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {{10}^9}\,\frac{{N{\rm{ - }}{m^2}}}{{{C^2}}}} \right]$

If the potential of the inner shell is $10\,V$ and that of the outer shell is $5\,V$, then potential at the centre will be....$V$

An electric charge $10^{-6} \mu \mathrm{C}$ is placed at origin $(0,0)$ $\mathrm{m}$ of $\mathrm{X}-\mathrm{Y}$ co-ordinate system. Two points $\mathrm{P}$ and $\mathrm{Q}$ are situated at $(\sqrt{3}, \sqrt{3}) \mathrm{m}$ and $(\sqrt{6}, 0) \mathrm{m}$ respectively. The potential difference between the points $P$ and $Q$ will be :

Assume that an electric field $\overrightarrow E  = 30{x^2}\hat i$ exists in space. Then the potential difference $V_A -V_O$, where $V_O$ is the potential at the origin and $V_A$ the potential at $x = 2\, m$ is

Two identical positive charges are placed on the $y$-axis at $y=-a$ and $y=+a$. The variation of $V$ (electric potential) along $x$-axis is shown by graph