First slide

Electric field

Question

Two semicircular rings lying in the same plane of uniform linear charge density $λ$ have radii r and 2r. 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 center of semi-circular rings is

Difficult

A

$\frac{1}{4 π ε_{o}} \frac{3 λ}{2 r}$
B

$\frac{1}{4 π ε_{o}} \frac{λ}{2 r}$
C

$\frac{1}{4 π ε_{o}} \frac{2 λ}{r}$
D

$\frac{1}{4 π ε_{o}} \frac{λ}{r}$

Solution

The electric field due to both straight wires shall cancel at common center O. The electric field due to larger and smaller semi-circular rings at O be E and E' respectively.
$A t c e n t e r o f A r c : E = \frac{2 k λ}{r} \sin (\frac{ϕ}{2}) F o r s e m i c i r c l e, t a k e ϕ = π \Rightarrow E = \frac{2 k λ}{r}$
$D u e t o L a r g e r S e m i c i r c u l a r r i n g : E = \frac{1}{4 π ε_{o}} \frac{2 λ}{2 r} D u e t o L a r g e r S e m i c i r c u l a r r i n g : E' = \frac{1}{4 π ε_{o}} \frac{2 λ}{r}$
$∴$ Magnitude of electric field at O is
$\Rightarrow E_{n e t} = \frac{1}{4 π ε_{o}} (\frac{2 λ}{r} - \frac{λ}{r}) = \frac{1}{4 π ε_{o}} \frac{λ}{r}$

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