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(A) The potential at any point $P$ due to a system of charges is given by $V = \sum \frac{1}{4 \pi \varepsilon_0} \frac{q_i}{r_i}$.In the isosceles tr

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(A) The potential at any point $P$ due to a system of charges is given by $V = \sum \frac{1}{4 \pi \varepsilon_0} \frac{q_i}{r_i}$.In the isosceles triangle $ABC$,charges $+q$ are placed at $A, B,$ and $C$. $D$ is the midpoint of $AB$ and $E$ is the midpoint of $AC$.By symmetry,the distance of $D$ from $A$ is $a$,from $B$ is $a$,and from $C$ is $\sqrt{AC^2 + AD^2 - 2(AC)(AD) \cos A} = \sqrt{(2a)^2 + a^2 - 2(2a)(a) \cos A}$.Similarly,the distance of $E$ from $A$ is $a$,from $C$ is $a$,and from $B$ is $\sqrt{AB^2 + AE^2 - 2(AB)(AE) \cos A} = \sqrt{(2a)^2 + a^2 - 2(2a)(a) \cos A}$.Since the configuration is symmetric with respect to the angle $A$,the potential at $D$ $(V_D)$ and the potential at $E$ $(V_E)$ are equal.$V_D = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{AD} + \frac{q}{BD} + \frac{q}{CD} \right) = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{a} + \frac{q}{a} + \frac{q}{CD} \right)$.$V_E = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{AE} + \frac{q}{CE} + \frac{q}{BE} \right) = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{a} + \frac{q}{a} + \frac{q}{BE} \right)$.Since $CD = BE$ by symmetry,$V_D = V_E$.The work done $W = Q(V_E - V_D) = Q(0) = 0$.

Three charges each of value $+q$ are placed at the corners of an isosceles triangle $ABC$ with sides $AB = AC = 2a$. The midpoints of $AB$ and $AC$ are $D$ and $E$ respectively. The work done in taking a charge $Q$ from $D$ to $E$ is (where $\varepsilon_0$ is the permittivity of free space):

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