Equal charges $q$ are placed at the vertices $A$ and $B$ of an equilateral triangle $ABC$ of side $a$. The magnitude of the electric field at point $C$ 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 $g=10 \ m/s^2$)

Two charges, each equal to $-q$, are kept at $(-a, 0)$ and $(a, 0)$. $A$ charge $q$ is placed at the origin. If $q$ is given a small displacement $y$ along the $y$-direction, the force acting on $q$ is proportional to:

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

$A$ conducting sphere of radius $10 \; cm$ has an unknown charge. If the electric field $20 \; cm$ from the centre of the sphere is $1.5 \times 10^{3} \; N/C$ and points radially inward,what is the net charge (in $nC$) on the sphere?

Electric lines of force about a negative point charge are: