$A$ uniformly charged semicircular arc of radius $r$ has linear charge density $\lambda$. What is the electric field at its centre? $(\epsilon_0 = \text{permittivity of free space})$

Two point charges $+8 q$ and $-2 q$ are located at $X=0$ (origin) and $X=L$ respectively. The net electric field due to these two charges is zero at point $P$ on the $X$-axis. The location of point $P$ from the origin is:

Suppose a uniformly charged wall provides a uniform electric field of $2 \times 10^4 \ N/C$ normally. $A$ charged particle of mass $2 \ g$ is suspended by a silk thread of length $20 \ cm$ and remains at a distance of $10 \ cm$ from the wall. Then the charge on the particle will be $\frac{1}{\sqrt{x}} \ \mu C$ where $x=$ . . . . . . . (Use $g=10 \ m/s^2$)

Four charges $q, 2q, -4q$ and $2q$ are placed in order at the four corners of a square of side $b$. The net field at the centre of the square is

$A$ ring of radius $R$ is charged uniformly with a charge $+Q$. The electric field at a point on its axis at a distance $r$ from any point on the ring will be

$A$ wire of length $L$ has a charge $Q$ distributed uniformly along its length. The wire is bent in the shape of a semicircle. The magnitude of the electric field at the centre of curvature of the semicircle is