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Two point charges q_1 = 2.10 nC and q_2 = -6.20 nC are 0.100 m apart. Point A is midway between them; point B is 0.080 m from q_1 and 0.060 m from q_2 (the figure (Figure 1) ). Take the electric potential to be zero at infinity.

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syed514

1)- Va= V(q1 on A)+V(q2 on A)= q1910^9/(d1) +q2910^9/(d2)..............if these particles in the air = 2.410^-9 910^9 /0.05 + -6.510^-9 910^9 / 0.05 = -738 volt. _ where : d1= the distance between q1 and A. d2 = the distance between q2 and A. 
2)- Vb= V(q1 on B)+V(q2 on A)= q1910^9/(d3)+ q2910^9/(d4) = 2.410^-9 910^9/0.08 + -6.510^-9 910^9/0.06 = -705volt 3)- work = qc(Va-Vb)= 2.810^-9 (-738-(-705))= -92.410^-9 J

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