MHT CET · Physics · Electrostatics
A uniformly charged semicircular arc of radius ' \(r\) ' has linear charge density ' \(\lambda\) '. The electric field at its centre is \(\left(\varepsilon_0=\right.\) permittivity of free space)
- A \(\frac{\lambda}{4 \varepsilon_0}\)
- B \(\frac{2 \varepsilon_0}{\lambda}\)
- C \(\frac{\lambda}{4 \varepsilon_0 r}\)
- D \(\frac{2 \pi \varepsilon_0}{\lambda}\)
Answer & Solution
Correct Answer
(C) \(\frac{\lambda}{4 \varepsilon_0 r}\)
Step-by-step Solution
Detailed explanation
\(\lambda =\frac{\mathrm{q}}{l} \)
\( \therefore \mathrm{q} =\lambda \times l=\lambda \times \pi \mathrm{r}\)
\(\therefore\) The electric field at its centre is,
\(
\begin{aligned}
& \mathrm{E}=\frac{\mathrm{q}}{4 \pi \varepsilon_0 \mathrm{r}^2}=\frac{\lambda \pi \mathrm{r}}{4 \pi \varepsilon_0 \mathrm{r}^2} \\
\therefore ~& \mathrm{E}=\frac{\lambda}{4 \varepsilon_0 \mathrm{r}}
\end{aligned}
\)
\( \therefore \mathrm{q} =\lambda \times l=\lambda \times \pi \mathrm{r}\)
\(\therefore\) The electric field at its centre is,
\(
\begin{aligned}
& \mathrm{E}=\frac{\mathrm{q}}{4 \pi \varepsilon_0 \mathrm{r}^2}=\frac{\lambda \pi \mathrm{r}}{4 \pi \varepsilon_0 \mathrm{r}^2} \\
\therefore ~& \mathrm{E}=\frac{\lambda}{4 \varepsilon_0 \mathrm{r}}
\end{aligned}
\)
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