A uniform electric field $E=3 \hat{i}+6 \hat{j}+\hat{k}$ passes through a closed cuboidal surface. One face of the cuboid has an area $4 m^2$ and an outward unit normal given by $\frac{2 \hat{i}+2 \hat{j}+3 \hat{k}}{\sqrt{17}}$. If the electric flux through the remaining 5 faces is zero, the charge enclosed by the cuboid is:

Point charge $\sqrt{2} C, \sqrt{2} C$, and $-2 C$ are placed at the three vertices of a right-angled triangle in air. [as shown in the figure below]

What is the electric field at a point $P$ on the hypotenuse that is equidistant from all three charges.

Given distances $X P=Y P=Z P=0.5 \mathrm{~m}$

A bob of a simple pendulum has a mass of 4 g and a charge of $20 \mu \mathrm{C}$. If it is at rest in a uniform horizontal electric field of intensity $1000 \mathrm{Vm}^{-1}$, the angle that the pendulum makes with the vertical at equilibrium is:

Three charges, $Q,-q$ and $2 q$ are placed at the vertices of a right-angled isosceles triangle. What is the value of $q$ for the net electrostatic energy of the configuration to be zero?