Equipotential Surfaces

Choose the incorrect statement. 

Equipotential surfaces 

If a charge $q$ is moved around a charge $Q$ in circular path of radius $r$ with $Q$ as centre then potential difference between any two points on circumference is

A non-uniform electric field is represented by equipotential lines. What is the direction of the electric field line at point $A$?

Equipotential spheres are drawn round a point charge. As we move away from the charge, will the spacing between two spheres having a constant potential difference decrease, increase or remain constant.

Referring to the spherical equipotential lines in Fig. $\left(\mathrm{a}\right)$ find

(a) $\overrightarrow{E}=f\left(r\right)$, (b) $\overrightarrow{E}$-pattern.

A positive charge q is moved from A to B in each diagram :

Draw the equipotential surface due to a system of two identical positive charges.

Draw the equipotential surface due to a dipole.

What will be the shape of equipotential surface in case of uniform electric field?

The shape of the equipotential surface due to a point charge is:

Assertion: Two equipotential surfaces cannot cut each other.
Reason: Two equipotential surfaces are parallel to each other.

Assertion(A): Electric field is always normal to equipotential surfaces and along the direction of decreasing order of potential.

Reason (R): Negative gradient of electric potential is electric field.

Nature of equipotential surface for a point charge is

What is equipotential surface? Draw the equipotential surface due to point charge.

Draw an equipotential surface in a uniform electric field.

What is the direction of the lines of force at any point on the equipotential surface?

The current following through a pure inductor of inductance $4 \mathrm{mH}$ is $i=12\cos 300t$ ampere. What is $rms$.

The instantaneous values of current and voltage in an circuit are given by $I=6\sin \left(100\mathrm{\pi t}+\frac{\mathrm{\pi }}{4}\right),V=5\sin \left(100\mathrm{\pi t}-\frac{\mathrm{\pi }}{4}\right)$. Then

In an ac circuit the current $B$ $i=5\sin \left(100t-\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$ $A$ and the A.C. potential is $V=200\sin \left(100t\right)$ Volt. Then the power consumed is.

Draw the equipotential surfaces due to an electric dipole. Locate the points where the potential due to the dipole is zero.