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

Step 1: Use symmetry, not brute force.
A regular hexagon has six vertices, all at the same distance $L$ from the centre. If all six held charge $+q$, the field at the centre would be exactly zero by symmetry.

Step 2: The missing-charge trick.
We have charges on only five vertices. Imagine adding a sixth $+q$ at the empty vertex. The full set gives zero field, so the five real charges produce a field equal and opposite to that one imagined charge.

Step 3: Field of one charge at the centre.
A single $+q$ at distance $L$ makes a field \[ E=\frac{1}{4\pi\varepsilon_0}\frac{q}{L^2}. \]

Step 4: Field due to the five charges.
By Step 2, the five charges give the same magnitude of field, \[ E_5=\frac{1}{4\pi\varepsilon_0}\frac{q}{L^2}. \]

Step 5: Force on the central charge.
A charge $-q$ at the centre feels $F=|{-q}|\,E_5$, so \[ F=q\cdot\frac{1}{4\pi\varepsilon_0}\frac{q}{L^2}. \]

Step 6: Final expression.
\[ F=\frac{1}{4\pi\varepsilon_0}\frac{q^2}{L^2}. \]
\[ \boxed{F=\dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q}{L}\right)^2} \]