Charges -q, q, q are placed at the vertices A | vedclass

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(C) The initial potential energy $U_i$ of the system is the sum of potential energies of the three pairs of charges:$U_i = \frac{1}{4\pi \epsilon_0} \

Vedclass · Accepted Answer

$\frac{q^2}{2\pi \epsilon_0 a}$

Vedclass · Answer

(C) The initial potential energy $U_i$ of the system is the sum of potential energies of the three pairs of charges:$U_i = \frac{1}{4\pi \epsilon_0} \left( \frac{(-q)(q)}{a} + \frac{(q)(q)}{a} + \frac{(q)(-q)}{a} \right) = \frac{1}{4\pi \epsilon_0} \left( -\frac{q^2}{a} + \frac{q^2}{a} - \frac{q^2}{a} \right) = -\frac{q^2}{4\pi \epsilon_0 a}$.When the charge $-q$ is at the midpoint $M$ of $BC$,the distance $AM = \sqrt{a^2 - (a/2)^2} = \frac{\sqrt{3}}{2}a$,and the distances $BM = MC = a/2$.The final potential energy $U_f$ is:$U_f = \frac{1}{4\pi \epsilon_0} \left( \frac{(-q)(q)}{a/2} + \frac{(-q)(q)}{a/2} + \frac{(q)(q)}{a} \right) = \frac{1}{4\pi \epsilon_0} \left( -\frac{2q^2}{a} - \frac{2q^2}{a} + \frac{q^2}{a} \right) = -\frac{3q^2}{4\pi \epsilon_0 a}$.By the law of conservation of energy,the change in kinetic energy $\Delta K$ is equal to the negative change in potential energy $-\Delta U$:$K_f - K_i = -(U_f - U_i) = U_i - U_f$.Since the charge starts from rest,$K_i = 0$.$K_f = -\frac{q^2}{4\pi \epsilon_0 a} - \left( -\frac{3q^2}{4\pi \epsilon_0 a} \right) = \frac{2q^2}{4\pi \epsilon_0 a} = \frac{q^2}{2\pi \epsilon_0 a}$.

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