Two point charges of magnitudes $-8 \mu C$ and $+32 \mu C$ are separated by a distance of $15 \ cm$ in air. The position of the point from the $-8 \mu C$ charge at which the resultant electric field becomes zero is (in $cm$):

The figure shows a rod $AB$,which is bent in a $120^{\circ}$ circular arc of radius $R$. $A$ charge $(-Q)$ is uniformly distributed over the rod $AB$. What is the electric field $\overrightarrow{E}$ at the centre of curvature $O$?

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:

For the given figure,the direction of the electric field at point $A$ will be:

The three charges $q/2, q$ and $q/2$ are placed at the corners $A, B$ and $C$ of a square of side '$a$' as shown in the figure. The magnitude of the electric field $(E)$ at the corner $D$ of the square is:

Two point charges $q_{A} = 3\; \mu C$ and $q_{B} = -3\; \mu C$ are located $20\; cm$ apart in vacuum.
$(a)$ What is the electric field at the midpoint $O$ of the line $AB$ joining the two charges?
$(b)$ If a negative test charge of magnitude $1.5 \times 10^{-9}\; C$ is placed at this point,what is the force experienced by the test charge?