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 ___________. (\( \varepsilon_0 \) is permittivity of free space) 

• A. \( \dfrac{Q}{4\pi \varepsilon_0 R^2}\left(\sqrt{3}\,\hat{i}-\hat{j}\right) \)
• B. \( -\dfrac{Q}{4\pi \varepsilon_0 R^2}\left(\sqrt{3}\,\hat{i}-\hat{j}\right) \)
• C. \( -\dfrac{5Q}{8\pi \varepsilon_0 R^2}\left(\hat{i}-3\hat{j}\right) \)
• D. \( -\dfrac{5Q}{8\pi \varepsilon_0 R^2}\left(\hat{i}+\sqrt{3}\hat{j}\right) \)

Hint:

In electric field problems with symmetric charge distributions, always resolve field vectors into components and use symmetry to cancel opposing contributions.

Answer 1

The Correct Option is B

Each charge is placed at a distance \( R \) from the center. The magnitude of electric field due to a charge \( Q \) at the center is \[ E = \frac{1}{4\pi \varepsilon_0}\frac{Q}{R^2}. \] Step 1: Analyze symmetry of charge distribution.
From the figure, the charges are placed at angular separations of \(60^\circ\). Due to symmetry, electric field vectors due to opposite charges partially cancel each other. Hence, we resolve the electric field vectors along the \(x\)-axis and \(y\)-axis.
Step 2: Resolve horizontal components.
Considering the directions of electric field due to each charge, the net horizontal component is proportional to \[ -\sqrt{3}. \] Step 3: Resolve vertical components.
Similarly, the net vertical component is proportional to \[ +1. \] Step 4: Write the resultant electric field vector.
Combining both components, \[ \vec{E} = -\frac{Q}{4\pi \varepsilon_0 R^2} \left(\sqrt{3}\,\hat{i}-\hat{j}\right). \] Final Answer: \[ \boxed{-\dfrac{Q}{4\pi \varepsilon_0 R^2} \left(\sqrt{3}\,\hat{i}-\hat{j}\right)} \]