Question

Find electric field intensity \( \vec{E} \) at the centre of the circle shown in the figure.

• A. \( \dfrac{KQ}{R^2}\hat{i} + \dfrac{KQ}{R^2}\hat{j} \)
• B. \( -\dfrac{\sqrt{3}KQ}{R^2}\hat{i} + \dfrac{KQ}{R^2}\hat{j} \)
• C. \( \dfrac{KQ}{R^2}\hat{i} + \dfrac{\sqrt{3}KQ}{R^2}\hat{j} \)
• D. \( \dfrac{\sqrt{3}KQ}{R^2}\hat{i} + \dfrac{\sqrt{3}KQ}{R^2}\hat{j} \)

Hint:

For symmetric charge distributions:
Always resolve fields into components
Use symmetry to cancel components wherever possible

Answer 1

The Correct Option is B

Concept:
Electric field due to a point charge at distance \(R\) is: \[ E = \frac{KQ}{R^2} \] The direction of the field is:
Away from a positive charge
Towards a negative charge The net electric field is the \emph{vector sum} of individual fields.

Step 1: Magnitude of field due to each charge. All charges are at the same distance \(R\) from the centre, hence each produces field of magnitude: \[ E_0 = \frac{KQ}{R^2} \]
Step 2: Resolve electric fields along \(x\) and \(y\) axes. From the figure:
Two \(+Q\) charges are symmetrically placed at angles \( \pm 30^\circ \) with the \(+x\)-axis
One \(+Q\) is at the bottom
One \(-Q\) is at the top-left Resolving components and summing: \[ E_x = -\sqrt{3}\,E_0 \] \[ E_y = +E_0 \]
Step 3: Write the resultant electric field. \[ \vec{E} = E_x\hat{i} + E_y\hat{j} \] \[ \vec{E} = -\frac{\sqrt{3}KQ}{R^2}\hat{i} + \frac{KQ}{R^2}\hat{j} \] \[ \boxed{\vec{E} = -\frac{\sqrt{3}KQ}{R^2}\hat{i} + \frac{KQ}{R^2}\hat{j}} \]