Question

Three charged particle A, B and C with charges -4q, 2q and -2q are present on the circumference of a circle of radius d. The charged particles A, C and centre O of the circle formed an equilateral triangle as shown in figure. Electric field at O along x-direction is:

• A. $\frac{2\sqrt{3}q}{\pi\epsilon_0d^2}$
• B. $\frac{\sqrt{3}q}{4\pi\epsilon_0d^2}$
• C. $\frac{3\sqrt{3}q}{4\pi\epsilon_0d^2}$
• D. $\frac{\sqrt{3}q}{\pi\epsilon_0d^2}$

Answer 1

Answer: (C)

$\frac{3\sqrt{3}q}{4\pi\epsilon_0d^2}$

Answer 2

Let O be the origin (0, 0). The radius of the circle is d. The distance of each charge from O is d.

The electric field at the origin O due to a charge q at position $\vec{r}$ is given by $\vec{E} = \frac{1}{4\pi\epsilon_0} \frac{q}{|\vec{r}|^2} (-\hat{r}) = \frac{1}{4\pi\epsilon_0} \frac{q}{d^2} (-\frac{\vec{r}}{d}) = -\frac{1}{4\pi\epsilon_0} \frac{q}{d^3} \vec{r}$.

The x-component of the electric field at O is $E_{Ox} = -\frac{1}{4\pi\epsilon_0d^2} (q_A \cos\theta_A + q_B \cos\theta_B + q_C \cos\theta_C)$.

Assuming OAC forms an equilateral triangle with side d and the x-axis is positioned symmetrically with respect to A and C. Let the angle of A be $-\alpha$ and the angle of C be $\alpha$. Then the angle between OA and OC is $2\alpha$. For $\angle AOC = 60^\circ$, we have $2\alpha = 60^\circ$, so $\alpha = 30^\circ$.

So, let $\theta_A = -30^\circ$ and $\theta_C = 30^\circ$. B is on the positive y-axis, so $\theta_B = 90^\circ$.

Now calculate the x-component of the electric field at O using these angles: $E_{Ox} = -\frac{1}{4\pi\epsilon_0d^2} ((-4q) \cos(-30^\circ) + (2q) \cos(90^\circ) + (-2q) \cos(30^\circ))$ $E_{Ox} = -\frac{q}{4\pi\epsilon_0d^2} (-4 \frac{\sqrt{3}}{2} + 2 \cdot 0 + (-2) \frac{\sqrt{3}}{2})$ $E_{Ox} = -\frac{q}{4\pi\epsilon_0d^2} (-2\sqrt{3} + 0 - \sqrt{3})$ $E_{Ox} = -\frac{q}{4\pi\epsilon_0d^2} (-3\sqrt{3})$ $E_{Ox} = \frac{3\sqrt{3}q}{4\pi\epsilon_0d^2}$.