The electric potential inside a conducting sphere:

$A$ bullet of mass $2\,g$ has a charge of $2\,\mu C$. Through what potential difference must it be accelerated, starting from rest, to acquire a speed of $10\,m/s$?

$A$ hollow metal sphere is charged with a charge of $3.2 \times 10^{-19} \ C$. If the radius of the sphere is $10 \ cm$,then the electric potential at a distance of $4 \ cm$ from its center will be:

$A$ regular hexagon of side $6 \ cm$ has a charge of $2 \ \mu C$ at each of its vertices. What is the potential at the centre of the hexagon? $\left[\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \text{ SI units}\right]$

$A$ charge $+q$ is placed at the origin $O$ of $X-Y$ axes as shown in the figure. The work done in taking a charge $Q$ from $A$ to $B$ along the straight line $AB$ is

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