Three point charges shown in the figure lie a | vedclass

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(C) The initial potential energy of the system is given by $U_i = \frac{1}{4 \pi \varepsilon_0} [\frac{(-q)(q)}{a} + \frac{(q)(-q)}{a} + \frac{(-q)(-q

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$\frac{q^2}{4 \pi \varepsilon_0 a}$

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(C) The initial potential energy of the system is given by $U_i = \frac{1}{4 \pi \varepsilon_0} [\frac{(-q)(q)}{a} + \frac{(q)(-q)}{a} + \frac{(-q)(-q)}{2a}] = \frac{1}{4 \pi \varepsilon_0} [-\frac{q^2}{a} - \frac{q^2}{a} + \frac{q^2}{2a}] = \frac{1}{4 \pi \varepsilon_0} [-\frac{3q^2}{2a}] = -\frac{3q^2}{8 \pi \varepsilon_0 a}$.After exchanging the central charge $+q$ with one of the negative charges $-q$,the new arrangement is $-q, -q, +q$ with distances $a$ between adjacent charges and $2a$ between the outer charges.The final potential energy is $U_f = \frac{1}{4 \pi \varepsilon_0} [\frac{(-q)(-q)}{a} + \frac{(-q)(q)}{a} + \frac{(-q)(q)}{2a}] = \frac{1}{4 \pi \varepsilon_0} [\frac{q^2}{a} - \frac{q^2}{a} - \frac{q^2}{2a}] = \frac{1}{4 \pi \varepsilon_0} [-\frac{q^2}{2a}] = -\frac{q^2}{8 \pi \varepsilon_0 a}$.The energy required is $\Delta U = U_f - U_i = -\frac{q^2}{8 \pi \varepsilon_0 a} - (-\frac{3q^2}{8 \pi \varepsilon_0 a}) = \frac{2q^2}{8 \pi \varepsilon_0 a} = \frac{q^2}{4 \pi \varepsilon_0 a}$.

Three point charges shown in the figure lie along a straight line. The energy required to exchange the position of the central charge with one of the negative charges is

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