Three point charges 1 C,-2 C,and -2 C are | vedclass

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(D) The potential energy of a system of point charges is given by $U = \sum \frac{k q_i q_j}{r_{ij}}$.For the initial state with side $a_1 = 1 \ m$:$U

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(D) The potential energy of a system of point charges is given by $U = \sum \frac{k q_i q_j}{r_{ij}}$.For the initial state with side $a_1 = 1 \ m$:$U_i = \frac{1}{4 \pi \varepsilon_0} [\frac{(1)(-2)}{1} + \frac{(-2)(-2)}{1} + \frac{(-2)(1)}{1}] = \frac{1}{4 \pi \varepsilon_0} [-2 + 4 - 2] = 0 \ J$.For the final state with side $a_2 = 2 \ m$:$U_f = \frac{1}{4 \pi \varepsilon_0} [\frac{(1)(-2)}{2} + \frac{(-2)(-2)}{2} + \frac{(-2)(1)}{2}] = \frac{1}{4 \pi \varepsilon_0} [-1 + 2 - 1] = 0 \ J$.The work done by an external force is $W_{ext} = U_f - U_i = 0 - 0 = 0 \ J$.

Three point charges $1 \ C$,$-2 \ C$,and $-2 \ C$ are placed at the vertices of an equilateral triangle of side $1 \ m$. The work done by an external force to increase the separation of the charges to $2 \ m$ in joules is $(\varepsilon_0 = \text{permittivity of air})$

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