Two semicircular rings lying in the same plane of uniform linear charge density λ have radii r and 2 r. They are joined using two straight uniformly charged wires of linear charge density λ and length r as shown in the figure. The magnitude of electric field at common centre of semicircular rings is

Question

Two semicircular rings lying in the same plane of uniform linear charge density λ have radii r and 2 r. They are joined using two straight uniformly charged wires of linear charge density λ and length r as shown in the figure. The magnitude of electric field at common centre of semicircular rings is <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/05c902c6a9dab2a4d6d460970f6556de-.png" /> (A) (1/4 π εo) (3 λ/2 r) (B) (1/4 π εo) (λ/2 r) (C) (1/4 π εo) (2 λ/r) (D) (1/4 π εo) (λ/r)

Tardigrade Admin · Accepted Answer

The electric field due to both straight wires shall cancel at common centre

O

.
The electric field due to larger and smaller semi-circular rings at

O

be

E

and

E^{'}

respectively.

E = \frac{1}{4 π ε _{o}} \frac{2 λ}{2 r} E^{'} \frac{1}{4 π ε _{o}} \frac{2 λ}{r}

∴

Magnitude of electric field at

O

is

= E - \frac{1}{4 π ε _{o}} (\frac{2 λ}{r} - \frac{λ}{r}) = \frac{1}{4 π ε _{o}} \frac{λ}{r}

$\frac{1}{4 π ε _{o}} \frac{3 λ}{2 r}$

$\frac{1}{4 π ε _{o}} \frac{λ}{2 r}$

$\frac{1}{4 π ε _{o}} \frac{2 λ}{r}$

$\frac{1}{4 π ε _{o}} \frac{λ}{r}$

4πεo​1​2r3λ​

4πεo​1​2rλ​

4πεo​1​r2λ​

4πεo​1​rλ​

$\frac{1}{4 π ε _{o}} \frac{3 λ}{2 r}$

$\frac{1}{4 π ε _{o}} \frac{λ}{2 r}$

$\frac{1}{4 π ε _{o}} \frac{2 λ}{r}$

$\frac{1}{4 π ε _{o}} \frac{λ}{r}$