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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)
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Verified Answer
The correct answer is:
$\frac{\lambda}{4 \varepsilon_0 r}$
$$
\begin{aligned}
\lambda & =\frac{\mathrm{q}}{l} \\
\therefore \quad \mathrm{q} & =\lambda \times l=\lambda \times \pi \mathrm{r}
\end{aligned}
$$
$\therefore \quad$ The electric field at its centre is,
$$
\begin{aligned}
& \mathrm{E}=\frac{\mathrm{q}}{4 \pi \varepsilon_0 \mathrm{r}^2}=\frac{\lambda \pi \mathrm{r}}{4 \pi \varepsilon_0 \mathrm{r}^2} \\
\therefore \quad & \mathrm{E}=\frac{\lambda}{4 \varepsilon_0 \mathrm{r}}
\end{aligned}
$$
\begin{aligned}
\lambda & =\frac{\mathrm{q}}{l} \\
\therefore \quad \mathrm{q} & =\lambda \times l=\lambda \times \pi \mathrm{r}
\end{aligned}
$$
$\therefore \quad$ The electric field at its centre is,
$$
\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 \quad & \mathrm{E}=\frac{\lambda}{4 \varepsilon_0 \mathrm{r}}
\end{aligned}
$$
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