Question

Deduce an expression for the electric field at a point on the equatorial plane of an electric dipole of length 2a

Answer 1

Explanation:

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Answer 2

Answer:

The expression for electric field,$E_p$ is $\frac{1}{4\pi\epsilon_0} \frac{Q}{d^3}$.

Explanation:

Let us consider an electric dipole $AB$ consisting of two charges $+q$ and $-q$ separated by the distance $2a$.

Also, let $P$ be the point on the equatorial line at a distance $d$ from mid-point $O$ of the dipole.

Now, the electric field, $E_1$ at point $P$ due to charge $-q$ is,

$E_1=\frac{1}{4\pi \epsilon_0 } \frac{q}{AP^2}$

⇒ $E_1=\frac{1}{4\pi \epsilon_0 } \frac{q}{(d^2+a^2)}$ . . . . . . (1)  (Since $AP^2=OA^2+OP^2$)

Similarly,

The electric field, $E_2$ at point $P$ due to charge $+q$ is,

$E_2=\frac{1}{4\pi \epsilon_0 } \frac{q}{PB^2}$

⇒ $E_2=\frac{1}{4\pi \epsilon_0 } \frac{q}{(d^2+a^2)}$ . . . . . . (2) (Since $PB^2=OB^2+OP^2$)

From (1) and (2), we get

$E_1=E_2$

Assume $E=E_1=E_2$.

Then the net electric field $E_p$ at $P$ due to the dipole is,

$E_p=E_1+E_2$

     $=E+E$

$E_p=2E$ . . . . . (3)

Now, from figure resolve components of $E_1$ as,

$E_1cos\theta$ and $E_1sin\theta$

Similarly, resolve components of $E_2$ as,

$E_2cos\theta$ and $E_2sin\theta$

From figure observe that $E_2sin\theta$ and $E_1sin\theta$ are equal and in opposite direction.

Thus, their resultant must be $0$.

Now,

$E_p=E_1cos\theta+E_2cos\theta$

    $=2Ecos\theta$ (As $E_1=E_2$)

Substitute the value of $E$,we get

$E_p=2\left( \frac{1}{4\pi \epsilon_0 } \frac{q}{(d^2+a^2)}\right)cos\theta$

$E_p=2\left( \frac{1}{4\pi \epsilon_0 } \frac{q}{(d^2+a^2)}\right)\frac{a}{d^2+a^2}$ (Since $cos\theta=\frac{a}{\sqrt{d^2+a^2} }$)

For a short dipole, $a < < d$

$E_p=\frac{1}{4\pi\epsilon_0} \frac{2aq}{d^3}$

Take $Q=2aq$,

⇒ $E_p=\frac{1}{4\pi\epsilon_0} \frac{Q}{d^3}$

Therefore, the expression for electric field,$E_p$ is $\frac{1}{4\pi\epsilon_0} \frac{Q}{d^3}$.

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