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Since the work done depends on the final arrangement of the charges, and not on how they are put together, we calculate work needed for one way of putting the charges at A, B, C and D. Suppose, first the charge +q is brought to A, and then the charges –q, +q, and –q are brought to B, C and D, respectively. The total work needed can be calculated in steps:

(i) Work needed to bring charge +q to A when no charge is present elsewhere: this is zero.

(ii) Work needed to bring –q to B when +q is at A.

This is given by (charge at B) × (electrostatic potential at B due to charge +q at A)

$\quad= -q \times \bigg( \large\frac{q}{4 \pi \in_0 d} \bigg) =-\large\frac{d^2}{4 \pi \in_0 d}$

(iii) Work needed to bring charge +q to C when +q is at A and –q is at B. This is given by (charge at C) × (potential at C due to charges at A and B)

$\quad= +q \bigg( \large\frac{+q}{4 \pi \in_0 d \sqrt 2} +\frac{-q}{4 \pi \in_0 d} \bigg)$

$\quad= \large\frac{-d^2}{4 \pi \in_0 d} \bigg( 1- \large\frac{1}{\sqrt 2} \bigg)$

(iv) Work needed to bring –q to D when +q at A,–q at B, and +q at C. This is given by (charge at D) × (potential at D due to charges at A, B and C)

$\quad= +q \bigg( \large\frac{+q}{4 \pi \in_0 d } +\frac{-q}{4 \pi \in_0 d \sqrt 2}+ \frac{q}{4 \pi \in_0d} \bigg)$

$\quad= \large\frac{-q^2}{4 \pi \in_0 d} \bigg( 2- \large\frac{1}{\sqrt 2} \bigg)$

Add the work done in steps (i), (ii), (iii) and (iv). The total work required is

$\quad= \large\frac{-q^2}{4 \pi \in_0 d} $$ \bigg( 4 -\sqrt 2 \bigg)$

The work done depends only on the arrangement of the charges, and not how they are assembled. By definition, this is the total electrostatic energy of the charges.

(We may try calculating same work/energy by taking charges in any other order desired and convince ourseleves that the energy will remain the same.)

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