A and B are two identical blocks made of a co | vedclass

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(C) At equilibrium,the electrostatic repulsive force between the two blocks is balanced by the spring force.The electrostatic force between the two bl

Vedclass · Accepted Answer

$2L\sqrt {4\pi {\varepsilon _0}K\left( {L - {L_0}} \right)} $

Vedclass · Answer

(C) At equilibrium,the electrostatic repulsive force between the two blocks is balanced by the spring force.The electrostatic force between the two blocks,each having charge $Q/2$,separated by a distance $L$ is given by Coulomb's Law:$F_e = \frac{1}{4 \pi \varepsilon_{0}} \frac{(Q/2)(Q/2)}{L^2} = \frac{1}{4 \pi \varepsilon_{0}} \frac{Q^2}{4L^2}$The spring force for an extension $(L - L_0)$ is given by Hooke's Law:$F_s = K(L - L_0)$Equating the two forces at equilibrium:$\frac{1}{4 \pi \varepsilon_{0}} \frac{Q^2}{4L^2} = K(L - L_0)$Solving for $Q^2$:$Q^2 = 4L^2 \cdot 4 \pi \varepsilon_{0} K(L - L_0)$$Q^2 = 16 \pi \varepsilon_{0} K L^2 (L - L_0)$Taking the square root on both sides:$Q = 4L \sqrt{\pi \varepsilon_{0} K (L - L_0)}$Wait,let us re-evaluate the provided options. If we simplify the expression:$Q^2 = 4 \pi \varepsilon_0 K (2L)^2 (L - L_0) / 1$ is not matching. Let's re-check the algebra:$\frac{Q^2}{16 \pi \varepsilon_0 L^2} = K(L - L_0)$$Q^2 = 16 \pi \varepsilon_0 L^2 K (L - L_0)$$Q = 4L \sqrt{\pi \varepsilon_0 K (L - L_0)}$Looking at option $C$: $2L \sqrt{4 \pi \varepsilon_0 K (L - L_0)} = 2L \cdot 2 \sqrt{\pi \varepsilon_0 K (L - L_0)} = 4L \sqrt{\pi \varepsilon_0 K (L - L_0)}$.Thus,option $C$ is correct.

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