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(a) Derive an expression for the electric field at any point on the equatorial line of an electric dipole.

(b) Two identical point charges, $q$ each, are kept $2m$ apart in air. A third point charge $Q$ of unknown magnitude and sign is placed on the line joining the charges such that the system remains in equilibrium. Find the position and nature of $Q$ .

Answers (1)

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(a) Let us consider the two charges $q$ and $-q$ , be seperated by a distance

$2a$ .

On the equatorial line of the dipole let any point $P$ .at a distance $r$ from the line joining the two charges.

Now,

$d=\sqrt{r^{2}+a^{2}}$

The electric field at $P$ due to the charge is : for $q$ , the electric field :

$E_{q}=\frac{1}{4\pi \epsilon _{0}}\frac{q}{d^{2}}=\frac{1}{4\pi \epsilon _{0}}\frac{q}{r^{2}+a^{2}}$

For $-q$ , the electric field :

$E_{-q}=\frac{1}{4\pi \epsilon _{0}}\frac{q}{d^{2}}=\frac{1}{4\pi \epsilon _{0}}\frac{q}{r^{2}+a^{2}}$

Adding horizontal components, such that vertical will cancel out.

Hence,

$\vec{E}=(\vec{E_{q}}+\vec{E_{-q}})i\cos \theta$

Let us suppose, $\hat{P}$ be the unit vector in the direction of the dipole moment.

$\vec{E}=(\vec{E_{q}}+\vec{E_{-q}})\cos \theta\hat{P}$

$E=2\; \frac{1}{4\pi \epsilon _{0}}\; \frac{q}{r^{2}+a^{2}}\cos \theta \hat{P}$

$\cos \theta =\frac{a}{d}=\frac{a}{\sqrt{r^{2}+a^{2}}}$

$\vec{E}=\frac{1}{4\pi \epsilon _{0}}\frac{2qa}{(r^{2}+a^{2})\frac{3}{2}}\hat{P}$

$\vec{E}=\frac{1}{4\pi \epsilon _{0}}\frac{2\vec{P}}{(r^{2}+a^{2})\frac{3}{2}}$

If, $r> > a$

Then, $E=\frac{1}{4\pi \epsilon _{0}}\frac{2\vec{P}}{r^{3}}$

(b) According to the Coulomb's law, All the charges must be in equilibrium and the force depends upon the distance and hence, the position of $Q$ must be in the mddle of the two charges.

Now, the nature of $Q$ must have opposite sign to that of $q$ , to balance the forces.

Posted by

rishi.raj

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