Question 883: Electrostats & Capacitance

Question

A uniformly charged semicircular arc of radius '$$r$$' has linear charge density $$(\lambda)$$, is the electric field at its centre? ( $$\in_0=$$ permittivity of free space)

Accepted Answer

The electric field at the centre due to a small element ds of the ring is given by
$$\mathrm{dE}=\frac{1}{4 \pi \epsilon_0} \frac{\lambda \mathrm{ds}}{\mathrm{r}^2}$$
The field can be resolved along $$\mathrm{X}$$ and $$\mathrm{Y}$$ directions as shown in the figure.
Similarly field due to symmetrically situated another element can also be resolved along $$\mathrm{X}$$ and $$\mathrm{Y}$$ directions. The X-components are in the same direction and get added while the Y-components are in opposite directions and cancel each other. Hence, the total field due to the range is
$$\begin{array}{rlrl}
\mathrm{E} & =\int \mathrm{dE} \cos 	heta \
	herefore \mathrm{E} & =\int \frac{1}{4 \pi \epsilon_0} \frac{\lambda d s}{r^2} \cos 	heta \
& =\int \frac{1}{4 \pi \epsilon_0} \frac{\lambda d 	heta \cos 	heta}{\mathrm{r}^2} \quad=\frac{1}{4 \pi \epsilon_0} \frac{\lambda}{\mathrm{r}} \int_{-\pi / 2}^{\pi / 2} \cos 	heta d 	heta \
& =\frac{1}{4 \pi \epsilon_0} \frac{\lambda}{\mathrm{r}}[\sin 	heta]_{0 \pi / 2}^{\pi / 2}  =\frac{1}{4 \pi \epsilon_0} \frac{\lambda}{\mathrm{r}}[1-(-1)] \
& =\frac{1}{4 \pi \epsilon_0} \frac{\lambda}{r} 2 \quad=\frac{\lambda}{2 \pi \epsilon_0 r}
\end{array}$$

Answer

$$\frac{\lambda}{4 \epsilon_0 r}$$

Answer

$$\frac{2 \pi \epsilon_0}{\lambda}$$

Answer

$$\frac{\lambda}{4 \epsilon_0}$$

Answer

$$\frac{2 \epsilon_0}{\lambda}$$

Electrostats & Capacitance

A uniformly charged semicircular arc of radius '$r' has linear charge density (\lambda), is the electric field at its centre? ( \in_0=$ permittivity of free space)

Choose the correct answer: