Two charges $+4e$ and $+e$ are at a distance $x$ apart. At what distance must a charge $q$ be placed from charge $+e$ so that it is in equilibrium?

$A$ 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

As shown in the figure,at what distance in $cm$ from point $A$ will the electric field be zero?

Two point charges $q_1$ and $q_2 (=q_1/2)$ are placed at points $A(0, 1)$ and $B(1, 0)$ as shown in the figure. The electric field vector at point $P(1, 1)$ makes an angle $\theta$ with the $x$-axis,then the angle $\theta$ is

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

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?