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>> $Two equal charges ‘q’ of opposite sign a$

Question

Two equal charges $‘ q ’$ of opposite sign are separated by a small distance $‘ 2 a ’$ . The electric intensity $‘ E ’$ at a point on the perpendicular bisector of the line joining the two charges at a very large distance $‘ r ’$ from the line is :

A

$\frac{1}{4 π ε _{0}} \frac{q a}{r ^{2}}$

B

$\frac{1}{4 π ε _{0}} \frac{2 q a}{r ^{3}}$

C

$\frac{1}{4 π ε _{0}} \frac{2 q a}{r ^{2}}$

D

$\frac{1}{4 π ε _{0}} \frac{q a}{r ^{3}}$

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Updated on : 2022-09-05

Solution

Verified by Toppr

Correct option is B)

Field in y direction gets canceled
as shown above
$E$ in x direction = $2 E c o s θ$
$= \frac{2 k q}{r ^{2}} \frac{a}{r} (c o s θ = \frac{a}{r})$
$= \frac{2 q a}{4 π ϵ _{0} r ^{3}}$
$∴ E = \frac{2 q a}{4 π ϵ _{0} r ^{3}} (\overset{x}{^})$

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