For the regular pentagon system shown in the | vedclass

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Question

(A) In a regular pentagon,if all vertices have equal charges,the net force at the center is zero due to symmetry.Let the charges at the five vertices

Vedclass · Accepted Answer

$\frac{KQq_0}{x^2}$

Vedclass · Answer

(A) In a regular pentagon,if all vertices have equal charges,the net force at the center is zero due to symmetry.Let the charges at the five vertices be $q_1, q_2, q_3, q_4, q_5$. Here,four vertices have charge $2Q$ and one vertex has charge $Q$.We can treat this as a system of five charges of $2Q$ each,plus an additional charge of $-Q$ at the top vertex.The net force due to five equal charges of $2Q$ at the center is zero.Therefore,the net force on $q_0$ is only due to the additional charge of $-Q$ at the top vertex.The force exerted by a charge $-Q$ at distance $x$ on $q_0$ is $F = \frac{K(-Q)q_0}{x^2}$.The magnitude of this force is $\frac{KQq_0}{x^2}$ directed towards the top vertex.

For the regular pentagon system shown in the figure,find the net electrostatic force on the charge $q_0$ placed at the center.

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