(a) Draw the equipotential surfaces due to an electric dipole. (b) Derive an expression for the electric field due to a dipole of dipole moment ⃗ p at a point on its perpendicular bisector.

Question

Philoid · Accepted Answer

(a)A pair of two charges equal in magnitude and opposite in signs separated by a very small distance is known as an electric dipole.

For equipotential surfaces where E is increasing uniformly d should decrease uniformly thatswhy the equipotential surfaces are compactly arranged closer to the charges and we can see that the electric field lines are always perpendicular to the equipotential surface.

(b) Solution:

we know that electric field is given by,

Here, r=

and

From the figure we can see that the sine components components get mutually cancelled and the cos components can be added.

Now we get,

Since magnitudes are same,

.......(1)

Now,

Substituting the values in equation 1 we get,

Now dipole moment =2qa

Now

So, a can be eliminated in the denominator.

This is the required equation and the direction of the electric field will be opposite to the direction of the electric dipole.

(a) Draw the equipotential surfaces due to an electric dipole.(b) Derive an expression for the electric field due to a dipole of dipole moment ⃗pat a point on its perpendicular bisector.

(a) Draw the equipotential surfaces due to an electric dipole.
(b) Derive an expression for the electric field due to a dipole of dipole moment ⃗pat a point on its perpendicular bisector.