Question 177: Electrostatics & Capacitance

Question

Six point charges are kept $60^{\circ}$ apart from each other on the circumference of a circle of radius $R$ as shown in figure. The net electric field at the center of the circle is $\_\_\_\_$ .
( $\epsilon_0$ is permittivity of free space)
![](https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1ml8uei1k/9e65ab98-b14c-49c4-8345-d42e155ebad2/1ce43580-023b-11f1-88dd-c7214e4b90e3/file-6y3zli1ml8uei1l.png?format=png)

Accepted Answer

The electric field $(\overrightarrow{\mathrm{E}})$ created by a point charge q at a distance r is given by Coulomb's Law:
$ |\overrightarrow{\mathrm{E}}|=\frac{1}{4 \pi \epsilon_0} \frac{|\mathrm{q}|}{\mathrm{r}^2} $
If the charge is positive $(+q)$, the electric field points away from the charge. If the charge is negative $(-q)$, the electric field points towards the charge.
When multiple charges are present, the net electric field at any point is simply the vector sum of the individual electric fields created by each charge.
Let the centre of the circle be the origin $(0,0)$. All six charges are at a distance R from the center.
Let the magnitude of the electric field produced by a single charge of magnitude $Q$ at the center as $E_0$ :
$ \mathrm{E}_0=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{Q}}{\mathrm{R}^2} $
The electric field due to charge +Q charge at $90^{\circ}$ is downward with magnitude $\mathrm{E}_0$ and that for the +Q at $270^{\circ}$ it is upward with magnitude $\mathrm{E}_0$. These two vectors are equal and opposite, so they cancel each other out.
The electric field due to charge +Q charge at $30^{\circ}$ is with magnitude $\mathrm{E}_0$ and opposite to that for the +Q at $210^{\circ}$ with magnitude $\mathrm{E}_0$. These two vectors are equal and opposite, so they cancel each other out.
$ \left|\overrightarrow{\mathrm{E}}_{n e t}\right|=\mathrm{E}_0+\mathrm{E}_0=2 \mathrm{E}_0 $
The direction of this net vector is $150^{\circ}$ or $30^{\circ}$ with negative x -axis.
So, the net electric field in vector form is,
$\overrightarrow{\mathrm{E}}_{\mathrm{net}}=\left(2 \mathrm{E}_0\right)\left(\cos 150^{\circ} \hat{\imath}+\sin 150^{\circ} \hat{\jmath}\right)$
$\Rightarrow $ $\overrightarrow{\mathrm{E}}_{\mathrm{net}}=\left(2 \mathrm{E}_0\right)\left(-\frac{\sqrt{3}}{2} \hat{\imath}+\frac{1}{2} \hat{\jmath}\right)$
$\Rightarrow $ $ \overrightarrow{\mathrm{E}}_{\mathrm{net}}=-\mathrm{E}_0(\sqrt{3} \hat{\imath}-\hat{\jmath}) $
Putting the expression for $\mathrm{E}_0$,
$ \overrightarrow{\mathrm{E}}_{\mathrm{net}}=-\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{Q}}{\mathrm{R}^2}(\sqrt{3} \hat{\imath}-\hat{\jmath}) $
Therefore, the net electric field at the centre of circle is $-\frac{1}{4 \pi \epsilon_0} \frac{Q}{R^2}(\sqrt{3} \hat{\imath}-\hat{\jmath})$.
Hence, the correct option is (C).
![](['https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1mlzdalxr/4033a8a0-1920-48ac-9a41-a50eebcdc016/5f0071f0-10d1-11f1-a982-f930035e5675/file-6y3zli1mlzdalxs.png?format=png', 'https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1mlzdcjte/a3fbd4fe-39bc-42f6-86a2-1b8dd72b5e5b/94fb4320-10d1-11f1-a982-f930035e5675/file-6y3zli1mlzdcjtf.png?format=png'])

Answer

$-\left(\frac{5 Q}{8 \pi \epsilon_0 R^2}\right)(\hat{i}-3 \hat{j})$

Answer

$\frac{Q}{4 \pi \in_{\mathrm{o}} R^2}(\sqrt{3} \hat{i}-\hat{j})$

Answer

$-\frac{\mathrm{Q}}{4 \pi \in_{\mathrm{o}} R^2}(\sqrt{3} \hat{i}-\hat{j})$

Answer

$-\frac{5 Q}{8 \pi \epsilon_{\mathrm{o}} R^2}(\hat{i}+\sqrt{3} \hat{j})$

Electrostatics & Capacitance

Six point charges are kept 60^{\circ} apart from each other on the circumference of a circle of radius R as shown in figure. The net electric field at the center of the circle is \_\_\_\_ . ( \epsilon_0 is permittivity of free space)

Choose the correct answer: