Assertion: Consider two identical charges placed at a distance $2d$ apart along the $x-$axis. The equilibrium of a positive test charge placed at the point $O$ midway between them is stable for displacements along the $x-$axis.
Reason: The net force on the test charge at point $O$ is zero.

$A$ cube of side $b$ has a charge $q$ at each of its vertices. The electric field due to this charge distribution at the centre of this cube will be

The earth's surface has a negative surface charge density of $10^{-9} \; C\;m^{-2}$. The potential difference of $400 \; kV$ between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only $1800 \; A$ over the entire globe. If there were no mechanism of sustaining the atmospheric electric field,how much time (roughly) would be required to neutralize the earth's surface (in $; s$)? (This never happens in practice because there is a mechanism to replenish electric charges,namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth $= 6.37 \times 10^{6} \; m$.)

Four charges are arranged at the corners of a square $ABCD$,as shown. The force on a $+ve$ charge kept at the centre of the square is

Draw the field lines for a bar magnet,a current-carrying finite solenoid,and an electric dipole.

$A$ solid sphere of radius $R$ carries a charge $(Q+q)$ distributed uniformly over its volume. $A$ very small point-like piece of it of mass $m$ gets detached from the bottom of the sphere and falls down vertically under gravity. This piece carries charge $q$. If it acquires a speed $v$ when it has fallen through a vertical height $y$ (see figure),then: (assume the remaining portion to be spherical).