First slide

Electric dipole

Question

Find the magnitude of the electric field at the point P in the configuration shown in figure for $d > > a$ . Take 2qa = p.

Difficult

A

$\frac{1}{4 {πε}_{0} d^{3}} \sqrt{q^{2} d^{2} + p^{2}}$
B

$\frac{1}{16 {πε}_{0} d^{3}} \sqrt{q^{2} d^{2} + p^{2}}$
C

$\frac{1}{2 {πε}_{0} d^{3}} \sqrt{q^{2} d^{2} + p^{2}}$
D

$\frac{1}{8 {πε}_{0} d^{3}} \sqrt{q^{2} d^{2} + p^{2}}$

Solution

The given situation is similar to the dipole at a point on an equatorial line (as shown in the following figure).From the figure, we can see that sin $θ$ components cancel out each other and cos $θ$ components add up. That is,
$E_{net} = 2 E (\cos θ) = 2 \times \frac{kq}{x^{2}} (\cos θ) = 2 \times \frac{1}{4 {πε}_{0}} [\frac{q}{{(a^{2} + d^{2})}^{3 / 2}}] a$
The dipole moment is given as p = 2qa. Substituting this value of p in the above relation, we get the magnitude of the net electric field at the point P (in the given configuration) due to dipole= $\frac{p}{4 {πε}_{o}} \frac{1}{{(a^{2} + d^{2})}^{3 / 2}}$
Now net electric field= $\sqrt{\frac{k^{2} q^{2}}{d^{4}} + \frac{k^{2} p^{2}}{{(a^{2} + d^{2})}^{3}}} = \sqrt{\frac{k^{2}}{d^{4}} (q^{2} + \frac{p^{2}}{d^{2}}}) = \sqrt{\frac{k^{2}}{d^{6}} (q^{2} d^{2} + p^{2})} = \frac{k}{d^{3}} \sqrt{q^{2} d^{2} + p^{2}}$
as $E_{net} = \frac{1}{4 {πε}_{0} d^{3}} \sqrt{q^{2} d^{2} + p^{2}}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App