The number of electrons to be put on a spherical conductor of radius $0.1\,m$ to produce an electric field of $0.036\,N/C$ just above its surface is:

The point charges $Q$ and $-2Q$ are placed at a distance $r$ apart. If the electric field at the location of $Q$ (point $A$) due to $-2Q$ is $\vec E$,then the electric field at the location of $-2Q$ (point $B$) due to $Q$ will be:

$A$ deuteron and an $\alpha$-particle are separated by a distance of $1\,\mathring{A}$ in air. The magnitude of the electric field due to the deuteron at the position of the $\alpha$-particle is:

$A$ cube of side $L$ has point charges $+q$ located at its seven vertices and $-q$ at remaining one vertex. The electric field at its centre is found to be $|E|=\alpha\left(\frac{q}{4 \pi \varepsilon_0 L^2}\right)$. The magnitude of constant $\alpha$ is

An electric field is given by $\vec{E} = e_1 \hat{i} + e_2 \hat{j} + e_3 \hat{k}$. $A$ charge $Q$ is moved by a displacement vector $\vec{r} = a \hat{i} + b \hat{j}$. The work done is:

The point charges $+q, -q, -q, +q, +Q$ and $-q$ are placed at the vertices $A, B, C, D, E$ and $F$ respectively of a regular hexagon $ABCDEF$. The electric field at the centre of the hexagon '$O$' due to the five charges at $A, B, C, D$ and $F$ is thrice the electric field at centre '$O$' due to charge $+Q$ at $E$ alone. The value of $Q$ is