$A$ regular hexagon of side $10 \; cm$ has a charge $5 \; \mu C$ at each of its vertices. Calculate the potential at the centre of the hexagon.

The electric charges $+2q$, $+2q$, $-2q$ and $-2q$ are placed at the corners of a square of side $2L$ as shown in the figure. The electric potential at point $A$, midway between the two charges $+2q$ and $+2q$, is $(\epsilon_0 = \text{permittivity of free space})$

$A$ charge $+q$ is fixed at each of the points $x = x_0, x = 3x_0, x = 5x_0, \dots$ up to $\infty$ on the $X$-axis and a charge $-q$ is fixed at each of the points $x = 2x_0, x = 4x_0, x = 6x_0, \dots$ up to $\infty$. Here $x_0$ is a positive constant. Taking the potential at a point due to a charge $Q$ at a distance $r$ from it to be $\frac{Q}{4\pi\varepsilon_0 r}$,find the potential at the origin due to the above system of charges.

$A$ charge of $5\,C$ is given a displacement of $0.5\,m$. The work done in the process is $10\,J$. The potential difference between the two points will be.......$V$

Let $V_1$ be the potential at the center of a square of side $1 \ m$ when the charges at the $4$ corners are $2 \ C$ each. If the same charges are placed at the corners of a square of side $2 \ m$,then the potential at the center of this square is $V_2$. The value of $\frac{V_2}{V_1}$ is

What is the change in kinetic energy of an $\alpha$-particle when it moves from a point at $70\ V$ to a point at $50\ V$?