Question

Derive the expression for field intensity due to 1
an electric dipole in vacuum for points on its axis.
Compare the variation of field intensity with
distance for an electric dipole and a point charge
when both are kept in vacuum.​

Answer 1

Answer:

Electric field on axis of dipole is

$E = \frac{2kP}{r^3}$

electric field of the point charge is given as

$E = \frac{kq}{r^2}$

so electric field of dipole decreases inversely with cube of the distance while electric field of the charge decrease inversely with square of the distance from charge

Explanation:

Let the dipole is consisting point charge "q" and having length L = 2a

now the electric field due to dipole at any point on its axis is given as

$E = \frac{kq}{(r - a)^2} + \frac{kq}{(r + a)^2}<strong>$

$E = kq(\frac{1}{(r - a)^2} + \frac{1}{(r + a)^2})$

$E = \frac{kq(4ra)}{(r^2 - a^2)^2}$

now let say the distance of dipole is large enough than its length

so we will have

$E = \frac{kq(4a)}{r^3}$

$E = \frac{2kP}{r^3}$

Now electric field of the point charge is given as

$E = \frac{kq}{r^2}$

so electric field of dipole decreases inversely with cube of the distance while electric field of the charge decrease inversely with square of the distance from charge

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Topic : Electric field due to dipole

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