Question

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

• A. $\frac{1}{4 \pi \varepsilon_{o}} \frac{3 \lambda}{2 r}$
• B. $\frac{1}{4 \pi \varepsilon_{o}} \frac{\lambda}{2 r}$
• C. $\frac{1}{4 \pi \varepsilon_{o}} \frac{2 \lambda}{r}$
• D. $\frac{1}{4 \pi \varepsilon_{o}} \frac{\lambda}{r}$

Answer 1

Answer: (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 $E^{\prime}$ respectively.
$E=\frac{1}{4 \pi \varepsilon_{o}} \frac{2 \lambda}{2 r} E^{\prime} \frac{1}{4 \pi \varepsilon_{o}} \frac{2 \lambda}{r}$
$\therefore $ Magnitude of electric field at $O$ is
$=E-\frac{1}{4 \pi \varepsilon_{o}}\left(\frac{2 \lambda}{r}-\frac{\lambda}{r}\right)=\frac{1}{4 \pi \varepsilon_{o}} \frac{\lambda}{r}$