5 charges each of magnitude 10^{-5} ,C and ma | vedclass

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(B) Initially,the net force on the central charge at $O$ is zero due to the symmetric arrangement of $5$ identical charges at $P_1, P_2, P_3, P_4,$ an

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$9 \,m/s^2$ downwards

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(B) Initially,the net force on the central charge at $O$ is zero due to the symmetric arrangement of $5$ identical charges at $P_1, P_2, P_3, P_4,$ and $P_5$.Let $\vec{F}_1, \vec{F}_2, \vec{F}_3, \vec{F}_4,$ and $\vec{F}_5$ be the forces exerted by the charges at $P_1, P_2, P_3, P_4,$ and $P_5$ respectively on the central charge at $O$.Since the net force is zero,$\vec{F}_1 + \vec{F}_2 + \vec{F}_3 + \vec{F}_4 + \vec{F}_5 = 0$.When the charge at $P_1$ is removed,the new net force $\vec{F}_{net}$ on the central charge is $\vec{F}_2 + \vec{F}_3 + \vec{F}_4 + \vec{F}_5 = -\vec{F}_1$.This means the net force is equal in magnitude to the force exerted by the charge at $P_1$ and directed opposite to it (i.e.,along the line $OP_1$ away from $P_1$,which is upwards).The magnitude of the force is $F = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2} = (9 	imes 10^9) \frac{(10^{-5})(5 	imes 10^{-5})}{(1)^2} = 4.5 \,N$.The acceleration $a$ of the central charge is $a = \frac{F}{m} = \frac{4.5 \,N}{0.5 \,kg} = 9 \,m/s^2$.Thus,the acceleration is $9 \,m/s^2$ upwards.

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