Question

Five point charges, each of value \(+q\), are placed on five vertices of a regular hexagon of side \(L\). What is the magnitude of the force on a point charge \(-q\) placed at the centre of the hexagon?

• A. \[ \frac{1}{4\pi\varepsilon_0} \left(\frac{q}{L}\right)^2 \]
• B. \[ \frac{1}{4\pi\varepsilon_0} \left(\frac{q}{L^2}\right) \]
• C. \[ \frac{1}{4\pi\varepsilon_0} \left(\frac{5q}{L}\right)^2 \]
• D. \[ \frac{1}{4\pi\varepsilon_0} \left(\frac{5q}{L^2}\right) \]

Hint:

For a regular hexagon, \[ R=L. \] When one vertex charge is missing, the resultant field of the remaining five charges equals the field due to the missing charge in magnitude.

Answer 1

The Correct Option is A

Concept: For a regular hexagon, \[ \text{distance from centre to each vertex}=L. \] If identical charges are placed at all six vertices, the net electric field at the centre is zero.

Step 1: Use the symmetry argument. Imagine a sixth charge \(+q\) is placed at the missing vertex. Then, \[ \vec E_1+\vec E_2+\cdots+\vec E_6=0. \] Hence, \[ \vec E_{\text{five charges}} = -\vec E_{\text{missing charge}}. \] Therefore the magnitude of the resultant electric field equals the field due to one charge.

Step 2: Calculate the field at the centre due to one vertex charge. \[ E = \frac{1}{4\pi\varepsilon_0} \frac{q}{L^2}. \]

Step 3: Find the force on the charge \(-q\). \[ F=qE \] \[ = q \left( \frac{1}{4\pi\varepsilon_0} \frac{q}{L^2} \right) \] \[ = \frac{1}{4\pi\varepsilon_0} \frac{q^2}{L^2}. \] \[\begin{aligned} \boxed{ F= \frac{1}{4\pi\varepsilon_0} \left(\frac{q}{L}\right)^2 } \end{aligned}\] Hence, option \(\mathbf{(A)}\) is correct.