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Derive the expression for the electric field of a dipole at a point on the equatorial plane of

the dipole

(b) Draw the equipotential surface due to an electric dipole. Locate the points where the

potential due to the dipole is zero.

Answer
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(a) Electric dipole moment- The strength of the electric dipole is measured by the quantity of electric dipole moment. Its magnitude equal to the product of the magnitude of their charge and the distance between the two charges

Electric dipole moment.

$ p = q \times d $

Electric dipole moment is a vector quantity and it is represented as $ \overrightarrow p = q \times \overrightarrow d $ in vector form.

Here p is the dipole moment, q is the charge and d is the distance and direction of $ \overrightarrow d $is from negative charge to positive charge.

Electric Field of dipole at points on the equatorial plane

Let the magnitude of the electric field due to the two charges +q and -q are given by $E_1$ and $E_2$ respectively.

Therefore from the figure, we have

\[AP = BP = \sqrt {{r^2} + {l^2}} \]

\[\therefore \overrightarrow {E_1} = \dfrac{1}{{4\pi \varepsilon o}}\dfrac{q}{{{r^2} + {l^2}}}\],

along B to P

\[\therefore \overrightarrow {E_2} = \dfrac{1}{{4\pi \varepsilon o}}\dfrac{q}{{{r^2} + {l^2}}}\],

along P to A

Hence both $E_1$ and $E_2$ are equal in magnitude.

Now, to find the resultant electric field, we have to resolve the components along and perpendicular to AB

The components perpendicular to AB are sine components and being equal and opposite to each other they cancel out each other.

Therefore the resultant electric field is given by,

$

E_1 = E_1\cos \theta + E_2\cos \theta \\

But, \\

E_1 = E_2 = \dfrac{1}{{4\pi \varepsilon o}}\dfrac{q}{{{r^2} + {l^2}}} \\

$

Thus clearly we can say that

$

\cos \theta = \dfrac{{OB}}{{PB}} = \dfrac{l}{{\sqrt {{r^2} + {l^2}} }} = \dfrac{l}{{{{({r^2} +

{l^2})}^{1/2}}}} \\

\therefore E = 2E_1\cos \theta \\

= 2 \times \dfrac{1}{{4\pi \varepsilon o}}\dfrac{q}{{({r^2} + {l^2})}}\dfrac{l}{{{{({r^2} +

{l^2})}^{1/2}}}} \\

= \dfrac{1}{{4\pi \varepsilon o}}\dfrac{{2ql}}{{{{({r^2} + {l^2})}^{3/2}}}} \\

$

If the dipole is infinitesimal and point P is far away then $ {l^2} $can be neglected as compared to $ {r^2} $

So,

$ E = \dfrac{1}{{4\pi \varepsilon o}}\dfrac{p}{{{r^3}}} $, parallel to $ \overrightarrow {BA} $

(b) Any surface over which the potential is constant is called an equipotential surface. In other words,

the potential difference between any two points on an equipotential surface is zero

Electric potential is zero at all points in the plane passing through the dipole equator.

Any surface over which the potential is constant is called an equipotential surface. In other words, the potential difference between any two points on an equipotential surface is zero.