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
Show Hint
Charges at opposite corners of the hexagon cancel or add in pairs at the centre.
Step 1: Understanding the Concept:
The six charges sit at the corners of a regular hexagon, all at the same distance \(r\) from the centre \(O\). The field of each point charge at \(O\) has magnitude \(\frac{kq}{r^2}\), pointing away from a positive charge and toward a negative charge.
Step 2: Key Formula or Approach:
From the figure: A \(= +q\), B \(= -q\), C \(= -q\), D \(= +q\), E \(= +Q\), F \(= -q\). Corners A and D are opposite each other, as are B and E, and C and F.
Step 3: Detailed Explanation:
A and D are both \(+q\) and opposite, so their fields at \(O\) are equal and opposite, and cancel.
C and F are both \(-q\) and opposite, so they cancel too.
Only B is left from the five charges. Its field is \(\frac{kq}{r^2}\), directed from \(O\) toward B.
The field of \(+Q\) at E alone is \(\frac{kQ}{r^2}\), directed away from E, which is also toward B.
Given: field of the five charges \(= 3 \times\) field of \(Q\) alone.
\[ \frac{kq}{r^2} = 3\cdot\frac{kQ}{r^2} \Rightarrow Q = \frac q3 \]
Final Answer:
The value of \(Q\) is \(\frac{q}{3}\), option (A).
\[ \boxed{\frac{q}{3}} \]
