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Three point charges 1C, 2C and 3C are placed at the corners of an equilateral triangle of side 1m.

The work required to move these charges to the corners of a smaller equilateral triangle of side 0.5 m in two differenct ways as in fig. (A) and fig. (B) are W_(a) and W_(b) then: 1
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Work done = Change in potential energy = U(final) - U(initial) , and,

U = k{ (Q1.Q2)/r1 +(Q2.Q3)/r2 +(Q3.Q1)/r3 }, where k = 9x10^9

Initially, when charges are kept in equilateral triangle of side 1 m :

Since triangle is equilateral, r1 = r2 = r3 = r = 1 m

Also, Q1 = 1C , Q2 = 2C, Q3 = 3C

Therefore,

U(initial) = k{1x2 + 2x3 + 3x1}/1 = 11k

Now

For position A :

All three charges are in a new equilateral triangle of side r = 0.5 m

Therefore,

U(final) for A = k{1x2 + 2x3 + 3x1}/0.5 = 22k

Hence, Work done in A = U(final) - U(initial) = 22k - 11k = 11k

For position B :

All three charges are in a new equilateral triangle of side r = 0.5 m

U(final) for B = k{1x2 + 2x3 + 3x1}/0.5 = 22k

hence, Work done in B = U(final) - U(initial) = 22k - 11k = 11k

From above, clearly, Work in A = Work in B

Hence, option C is true.

Above was the complete approach, but since the options do not deal with exact data, we can use another approach which will be much shorter for this question.

As we know, Work done of a conservative force is path independent, i.e, it only depends on initial and final conditions of the system.

In our question Coulomb's Force is mentioned, which is a conservative force, hence we will just check for initial and final conditions.

Initially, system is an equilateral triangle of side 1 m. Finally, system is an equilateral triangle of side 0.5 m for both A and B. Hence, as these conditions are same, the work done will also be the same, irrespective of the path followed to achieve the final condition in method A or method B. thus,

Work in A = Work in B