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(A) In a regular hexagon, the distance from the center to any vertex is equal to the side length $L$. Let the vertices be $V_1, V_2, V_3, V_4, V_5, V_

Vedclass · Accepted Answer

$k\frac{Q^2}{L^2}$

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(A) In a regular hexagon, the distance from the center to any vertex is equal to the side length $L$. Let the vertices be $V_1, V_2, V_3, V_4, V_5, V_6$. Suppose charges $+Q$ are placed at $V_1, V_2, V_3, V_4, V_5$ and the center $O$ has a charge $-Q$.If there were a charge $+Q$ at $V_6$ as well, the net force on the center $O$ would be zero due to symmetry (as each pair of opposite charges would exert equal and opposite forces).Let $\vec{F}_1, \vec{F}_2, \vec{F}_3, \vec{F}_4, \vec{F}_5$ be the forces exerted by the charges at the five vertices on the center charge $-Q$. Let $\vec{F}_6$ be the force that would be exerted by a charge $+Q$ at $V_6$.By symmetry, $\vec{F}_1 + \vec{F}_2 + \vec{F}_3 + \vec{F}_4 + \vec{F}_5 + \vec{F}_6 = 0$.Therefore, the net force $\vec{F}_{net} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 + \vec{F}_4 + \vec{F}_5 = -\vec{F}_6$.The force $\vec{F}_6$ exerted by a charge $+Q$ at $V_6$ on $-Q$ at the center is attractive and has a magnitude $F = k\frac{|(+Q)(-Q)|}{L^2} = k\frac{Q^2}{L^2}$.Since $\vec{F}_{net} = -\vec{F}_6$, the magnitude of the net force is $|\vec{F}_{net}| = |-\vec{F}_6| = k\frac{Q^2}{L^2}$.

Five charges, each $+Q$, are placed at five vertices of a regular hexagon of side length $L$. What is the magnitude of the force on a charge $-Q$ placed at the center of the hexagon?

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