Question

A dipole of length 0.1 m consists of two charges of +- 500 microcoulomb. What is the electric field due to the dipole at a point on the axis distant 0.2 m from the centre of the dipole.​

Answer 1

Electric field = 2kp/r^3

Answer 2

Concept:

The product of either charge and the separation distance (d) between them produces the dipole moment magnitude. It is defined by the formula, $p=2ql$

Given:

Length of dipole (2l) = 0.1 m

Charge on dipole = 500 μC = $500$ × $10^{-6}$ C

Distance of point p from centre of the dipole = 0.2 m

Find:

We are asked to determine the dipole Moment and electric field due to the dipole at a point on the axis 0.2 m from the centre of the dipole.

Solution:

According to the formula of the dipole moment, $p=2ql$

The dipole moment becomes, $p=500$ × $10^{-6}$ × $0.1$

Therefore, dipole moment, $p=5$ × $10^{-5} Cm$.

Distance of point p from the centre of the dipole, r = 0.2 m

Thus, the electric field at the point P becomes, $E=\frac{k2pr}{[r^{2}-l^{2}]^{2} }$

Where, $k= \frac{1}{4}$πε₀ = $9$ × $10^{9}$

Electric Field becomes, $E=\frac{9*10^{9} *2*5*10^{-5}0.2 }{[(0.2)^{2}-0.05^{2}]^{2} }$

Therefore, $E=\frac{18*10^{4} }{0.00140625}$

The electric field becomes, $E=1.28$ × $10^{8} NC^{-1}$

Thus, the electric dipole moment is $5$×$10^{-5} Cm$ and Electric Field is $1.28$ ×$10^{8} NC^{-1}$.

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