The magnitude of electric field intensity $E$ such that an electron placed in it would experience an electrical force equal to its weight is given by:

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:

$A$ circular wire loop of radius $1 \, cm$ carries a total charge $1 \times 10^{-6} \, C$ distributed uniformly over its length. If $0.01 \%$ of its length (circumference) is cut off, then the electric field at the centre of the loop due to the remaining wire is
$(\text{Take} \, \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \, \text{SI unit})$

$A$ point charge of $0.009 \ \mu C$ is placed at the origin. Calculate the electric field intensity due to this point charge at the point $(\sqrt{2}, \sqrt{7}, 0)$.

Why is an electric force conservative?

$A$ semi-infinite non-conducting rod lies along the $+x$-axis with its left end at the origin. The rod has a uniform linear charge density $\lambda$. The magnitude of the electric field $|\vec{E}|$ at a point on the $y$-axis at a distance $L$ from the origin will be: