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Question

Physics Question on Electric Charge

A charge qq is uniformly distributed on a ring of radius rr. A sphere of an equal radius is constructed with its centre lying on the periphery of the ring. The flux of electric field through the surface of the sphere will be

A

qε0\frac{q}{\varepsilon_{0}}

B

q2ε0\frac{q}{2\varepsilon_{0}}

C

q3ε0\frac{q}{3\varepsilon_{0}}

D

q4ε0\frac{q}{4\varepsilon_{0}}

Answer

q3ε0\frac{q}{3\varepsilon_{0}}

Explanation

Solution

Charge on ring qq, centre of ring =O=O Centre of sphere =O=O' Linear charge density of ring, λ=q2πa\lambda=\frac{q}{2 \pi a} Charge on arc ABA B of ring, qAB=λ(arcAB)=12πaa2π3q_{A B}=\lambda(\operatorname{arc} A B)=\frac{1}{2 \pi a} \cdot a \cdot \frac{2 \pi}{3} qAB=q/3q_{A B}=q / 3 i.e.., charged enclosed by sphere =q/3=q / 3 \therefore Flux coming out of sphere =q/3ε0=q / 3 \varepsilon_{0}