Question 841: Electrostats & Capacitance

Question

A uniformly charged semicircular arc of radius '$$r$$' has linear charge density '$$\lambda$$'. The electric field at its centre is ( $$\varepsilon_0=$$ permittivity of free space)

Accepted Answer

$$\begin{aligned}
\lambda & =\frac{\mathrm{q}}{l} \
	herefore \quad \mathrm{q} & =\lambda 	imes l=\lambda 	imes \pi \mathrm{r}
\end{aligned}$$
$$	herefore \quad$$ The electric field at its centre is,
$$\begin{aligned}
\quad \mathrm{E} & =\frac{\mathrm{q}}{4 \pi \varepsilon_0 \mathrm{r}^2}=\frac{\lambda \pi \mathrm{r}}{4 \pi \varepsilon_0 \mathrm{r}^2} \
	herefore \quad \mathrm{E} & =\frac{\lambda}{4 \varepsilon_0 \mathrm{r}}
\end{aligned}$$

Answer

$$\frac{\lambda}{4 \varepsilon_0}$$

Answer

$$\frac{2 \varepsilon_0}{\lambda}$$

Answer

$$\frac{\lambda}{4 \varepsilon_0 \mathrm{r}}$$

Answer

$$\frac{2 \pi \varepsilon_0}{\lambda}$$

Electrostats & Capacitance

A uniformly charged semicircular arc of radius '$r' has linear charge density '\lambda'. The electric field at its centre is ( \varepsilon_0=$ permittivity of free space)

Choose the correct answer: