Electrostatics 1 Question 19

20. Five point charges, each of value $+q$ coulomb, are placed on five vertices of a regular hexagon of side $L$ metre. The magnitude of the force on the point charge of value $-q$ coulomb placed at the centre of the hexagon is ……….newton.

(1992, 1 M)

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Answer:

Correct Answer: 20. $9 \times 10^{9} \frac{q^{2}}{L^{2}}$

Solution:

  1. Force on $-q$ due to charges at 1 and 4 are equal and opposite. Similarly, forces on $-q$ due to charges at 2 and 5 are also equal and opposite. Therefore, net force on $-q$ due to charges at 1, 2, 4 and 5 is zero. Only unbalanced force is between $-q$ and $+q$ at 3 which is equal to

$$ \begin{gathered} \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q^{2}}{L^{2}} \\ \text { or } \quad 9.0 \times 10^{9} \frac{q^{2}}{L^{2}} \text { (attraction) } \end{gathered} $$



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