The electric field due to a point charge $2q$ at a distance $r$ is $E$. Now,if charge $q$ is uniformly distributed over a thin spherical shell of radius $R$,the electric field at a distance $\frac{r}{2}$ $(r \gg R)$ from the center of the thin spherical shell is $E'=$ . . . . . . .

Point charges $+8q$ and $-2q$ are placed at $x = 0$ and $x = L$ respectively. At which point on the $X$-axis will the resultant electric field intensity be zero?

$A$ positively charged ball hangs from a silk thread. We put a positive test charge $q_0$ at a point and measure $F/q_0$. Then, it can be predicted that the electric field strength $E$ is:

The charges $2q, -q, -q$ are located at the vertices of an equilateral triangle. At the center of the triangle:

Four point positive charges of same magnitude $(Q)$ are placed at four corners of a rigid square frame of side $L$ as shown in the figure. The plane of the frame is perpendicular to the $Z$-axis. If a negative point charge $(-q)$ is placed at a small distance $z$ away from the center of the frame along the $Z$-axis $(z < < L)$, then:

Charge $q$ is uniformly distributed over a thin half ring of radius $R$. The electric field at the centre of the ring is