Question

A ring of radius r is marked in six equal parts and these parts are charged uniformly with a charge of magnitude q but positive and negative alternately as shown. Then the electric field at centre of ring will be –

Answer 1

Answer:

strong due to presence of negative and positive charge

Explanation:

hope you got it so plse mark as brainlist

Answer 2

Answer:

The electric field at the centre of the ring will be $\frac{1}{4 \pi \varepsilon_{0}} \frac{2 Q}{\pi R^{2}}$

Explanation:

We take an element on the ring.

Field at centre due to this element of charge

$&\mathrm{dE}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{dq}}{\mathrm{R}^{2}}\\=\frac{1}{4 \pi \varepsilon_{0}} \frac{\lambda \mathrm{Rd} \theta}{\mathrm{R}^{2}}\\=\frac{\lambda \mathrm{d} \theta}{4 \pi \varepsilon_{0} \mathrm{R}}\\=\frac{\mathrm{Q} \cdot \mathrm{d} \theta}{\pi \mathrm{R}\left(4 \pi \varepsilon_{0} \mathrm{R}\right)}\\=\frac{\mathrm{Q} d \theta}{4 \pi^{2} \varepsilon_{0} \mathrm{R}^{2}}$

$&\mathrm{E}_{\mathrm{x}}=\int \mathrm{dE}_{\mathrm{x}}\\=\int \mathrm{dE} \cos \theta\\=\int_{-\pi / 2}^{+\pi / 2} \frac{\mathrm{Q}}{4 \pi^{2} \varepsilon_{0} \mathrm{R}^{2}} \cos \theta \mathrm{d} \theta\\=\frac{\mathrm{Q}}{4 \pi^{2} \varepsilon_{0} \mathrm{R}^{2}}[\sin \theta]_{-\pi / 2}^{\pi / 2}\\=\frac{\mathrm{Q}}{2 \pi^{2} \varepsilon_{0} \mathrm{R}^{2}}$

$&\mathrm{E}_{\mathrm{y}}\\=\int \mathrm{dE}_{\mathrm{y}}\\=\int \mathrm{dE} \sin \theta\\=\frac{\mathrm{Q}}{4 \pi^{2} \varepsilon_{0} \mathrm{R}^{2}} \cdot \int_{-\pi / 2}^{\pi / 2} \sin \theta \mathrm{d} \theta$

$&=\frac{\mathrm{Q}}{4 \pi^{2} \varepsilon_{0} \mathrm{R}^{2}} \cdot(-\cos \theta)_{\pi / 2}^{\pi / 2}=0$

$&\mathrm{E}=\mathrm{E}_{\mathrm{x}}=\frac{\mathrm{Q}}{2 \pi^{2} \varepsilon_{0} \mathrm{R}^{2}} \text { along the axis of half ring. }\end{aligned}$