Electrostatics Question 661
Two semicircular rings lying in the same plane with uniform linear charge density $ \lambda $ have radii r and 2r. They are joined using two straight uniformly charged wires of linear charge density $ \lambda $ and length r. The magnitude of electric field at common centre of semi-circular rings is
Options:
A) $ \frac{1}{4\pi {\varepsilon _{0}}}\frac{3\lambda }{2r} $
B) $ \frac{1}{4\pi {\varepsilon _{0}}}\frac{\lambda }{2r} $
C) $ \frac{1}{4\pi {\varepsilon _{0}}}\frac{2\lambda }{r} $
D) $ \frac{1}{4\pi {\varepsilon _{0}}}\frac{\lambda }{r} $
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Answer:
Correct Answer: D
Solution:
[d] 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\pi {\varepsilon _{0}}}\frac{2\lambda }{2r}E’\frac{1}{4\pi {\varepsilon _{0}}}\frac{2\lambda }{r} $
$ \therefore $ Magnitude of electric field at O is $ =E-\frac{1}{4\pi {\varepsilon _{0}}}( \frac{2\lambda }{r}-\frac{\lambda }{r} )=\frac{1}{4\pi {\varepsilon _{0}}r}\lambda $
BETA
