Obtain an expression for the electric field intenstiy at a point on the equatorial line of an electric dipole.Solution in Kannada

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Let q be the magnitude of charge on the dipole. Let 2a be the dipole length. Let r be the distance of a point from the centre of the dipole. Let PO be perpendicular to AB. The directin of ${\to E}_{- q}$ and ${\to E}_{+ q}$ are as shown in the fig.
Resolving ${\to E}_{q}$ and ${\to E}_{- q}$ along two perpendicular directions, we not the $sin θ$ components of $E_{+ q}$ and $E_{- q}$ cancel each other and $cos θ$ components reinforce each other at that point.
Sicne $∣ ∣ {\to E}_{+ q} ∣ ∣ = ∣ ∣ {\to E}_{- q} ∣ ∣$ and ${\to E}_{B} = 2 E cos θ (- ˆ p)$
$∴ E_{B} = 2 E cos θ$
i.e. $E_{b} = 2 = (\frac{1}{4 π ε_{0}}) (\frac{q}{A P^{2}}) (\frac{a}{A P})$
where $A P = {(r^{2} + a^{2})}^{\frac{1}{2}}$
$E_{B} = (\frac{p}{4 π ε_{0}}) ⎛ ⎜ ⎝ \frac{q}{{(r^{2} + a^{2})}^{\frac{3}{2}}} ⎞ ⎟ ⎠$
where $p = 2 q a =$ dipole moment.
Since the direction of $E_{B}$ is opposite to $\to p$
${\to E}_{B} = (\frac{p}{4 p ε_{0}}) ⎛ ⎜ ⎝ \frac{1}{{(r^{2} + q^{2})}^{\frac{3}{2}}} ⎞ ⎟ ⎠ ˆ p$
where $p ˆ p = \to p$
${\to E}_{B} = \frac{- \to p}{(4 π ε_{0}) {(r^{2} + a^{2})}^{3 / 2}}$

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Updated on:21/07/2023

Class 12 PHYSICS ELECTROSTATICS

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