The electric field in a certain region is acting radially outward and is given by E = Ar. A charge contained in a sphere of radius 'a' centered at the origin of the field will be given by:

Two identical charged spheres suspended from a common point by two massless strings of lengths l are initially at a distance d (d << l) apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity v. Then v varies as a function of the distance x between the spheres, as:

Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude $17.0 \times 10^{-22} \, \text{C/m}^2$. The electric field between the plates is:

Two point charges $q_A = 3 \, \mu\text{C}$ and $q_B = -3 \, \mu\text{C}$ are located $20 \, \text{cm}$ apart in a vacuum. The electric field at the midpoint $O$ of the line $AB$ joining the two charges is:

The electric field due to a uniformly charged solid sphere of radius R as a function of the distance from its centre is represented graphically by -

Four-point +ve charges of the same magnitude (Q) are placed at four corners of a rigid square frame as shown in the figure. The plane of the frame is perpendicular to Z-axis. If a –ve point charge is placed at a distance z away from the above frame (z<<L) then

Shown below is a distribution of charges. The flux of electric field due to these charges through the surface S is:

Two point charges A and B, having charges +Q and −Q respectively, are placed at a certain distance apart and the force acting between them is F. If 25% charge of A is transferred to B, then the force between the charges becomes:

What is the flux of electric field $\vec{E} = 3 \times 10^3 \hat{i}$ N/C through a square of 10 cm on a side whose plane is parallel to the yz-plane?