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
$\frac{1}{{4\pi {\varepsilon _0}}}.\frac{{2q}}{{\sqrt {{a^2} + {b^2}} }}$
Zero
$\frac{1}{{4\pi {\varepsilon _0}}}.\frac{q}{{\sqrt {{a^2} + {b^2}} }}$
$\frac{1}{{4\pi {\varepsilon _0}}}.\frac{{( - q)}}{{\sqrt {{a^2} + {b^2}} }}$
In a region, if electric field is defined as $\vec E = \left( {\hat i + 2\hat j + \hat k} \right)\,V/m$ , then the potential difference between two points $A (0, 0, 0)$ and $B (2, 3, 4)$ in that region, is ......$V$
Consider two charged metallic spheres $S_{1}$ and $\mathrm{S}_{2}$ of radii $\mathrm{R}_{1}$ and $\mathrm{R}_{2},$ respectively. The electric $\left.\text { fields }\left.\mathrm{E}_{1} \text { (on } \mathrm{S}_{1}\right) \text { and } \mathrm{E}_{2} \text { (on } \mathrm{S}_{2}\right)$ on their surfaces are such that $\mathrm{E}_{1} / \mathrm{E}_{2}=\mathrm{R}_{1} / \mathrm{R}_{2} .$ Then the ratio $\left.\mathrm{V}_{1}\left(\mathrm{on}\; \mathrm{S}_{1}\right) / \mathrm{V}_{2} \text { (on } \mathrm{S}_{2}\right)$ of the electrostatic potentials on each sphere is
At distance of $5$ $cm$ and $10$ $cm $ outwards from the surface of a uniformly charged solid sphere, the potentials are $100$ $V$ and $75$ $V$ respectively . Then
A charge $Q$ is distributed over three concentric spherical shell of radii $a, b, c (a < b < c)$ such that their surface charge densities are equal to one another. The total potential at a point at distance $r$ from their common centre, where $r < a$, would be
A uniform electric field of $20\, N/C$ exists along the $x$ -axis in a space. The potential difference $(V_B -V_A)$ for the point $A(4\,m, 2\,m)$ and $B(6\,m, 5\,m)$ is.....$V$