Electrostatics Ques 18
- 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)
Show Answer
Answer:
Correct Answer: 18.$9 \times 10^{9} \frac{q^{2}}{L^{2}}$
Solution:
Formula:
Coulomb Force Between Two Point Charges:
- 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
$$ \begin{gathered} \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q^{2}}{L^{2}} \\ \text { or } \quad 9.0 \times 10^{9} \frac{q^{2}}{L^{2}} \text { (attraction) } \end{gathered} $$
BETA
