A 26 C point charge is placed 54 cm from an identical 26 C p

A +26 C point charge is placed 5.4 cm from an identical +26 C point charge.

How much work would be required by an external force to move a +0.12 C test charge from a point midway between them to a point 1.5 cm closer to either of the charges?

Express your answer using two significant figures.

(answer is not .51, 2.6, or 2.1)

Solution

the electric potential is,

     V = kq/r

the work done is calculated as follows:
    W = qV

         = q (Vb - Va)

         = q[kQ/rb - kQ/ra]

         = kqQ[1/rb - 1/ra]

         = (9x10⁹){0.12x10^-6}(26x10^-6) [(1/{[(0.054/2) -0.015]} + 1/{[(0.054/2) +0.015]}) - (1/{0.054/2} + 1/{0.054/2})]

         = 0.93 J