Question

The point charges \(+q\), \(-q\), \(-q\), \(+q\), \(+Q\) and \(-q\) are placed at the vertices of a regular hexagon ABCDEF as shown in the figure.

The electric field at the centre of the hexagon 'O' due to the five charges at A, B, C, D and F is thrice the electric field at centre 'O' due to charge \(+Q\) at E alone. The value of Q is

• A. \(\frac{q}{3}\)
• B. \(\frac{q}{4}\)
• C. \(3q\)
• D. \(4q\)

Hint:

Charges at opposite corners of the hexagon cancel or add in pairs at the centre.

Answer 1

The Correct Option is A

Step 1: Pair up opposite corners:
Opposite equal charges give zero net field at $O$ when they have the same sign: (A, D) and (C, F) both cancel.

Step 2: Compare the remaining charges:
What remains of the five is the single charge B $= -q$, giving a field $E_B = \frac{kq}{r^2}$ along $OB$. The sixth charge E gives $E_E = \frac{kQ}{r^2}$, also along $OB$.
Since $E_B = 3E_E$, we get $q = 3Q$, so $Q = \frac q3$.

Final Answer:
$Q = \frac{q}{3}$, option (A). \[ \boxed{\frac{q}{3}} \]