(b) At what distance from charge +Q1 on the line joining the two charges (in terms of Q1, Q2 and r) will this work done be zero.
[2020 | Delhi | Section - C | SET - 3, Q.32]
(a) Work done in placing the charge Q3 at the mid point
= Q3×net potential at the mid point
= Q3{k[Q1/(r/2)]+k[-Q2/(r/2)]}
W = 2kQ3[Q1-Q2]/r
b) Let the required distance be x,
Case i) When the required point will be on the left of Q1.
Work done at this point to be zero, when
0 = W1+W2
0 = kQ1Q3/x + k(-Q2)(Q3)/(r+x)
0 = kQ1Q3/x-kQ2Q3/(r+x)
-kQ1Q3/x = -kQ2Q3/(r+x)
Q1/x = Q2/(r+x)
Q1r+Q1x = Q2x
Q1r = Q2x-Q1x
Q1r = (Q2-Q1)x
Q1r/(Q2-Q1) = x
x = Q1r/(Q2-Q1)
x = -Q1r/(Q1-Q2)
Case ii) When the required point will be on the right of -Q2.
Work done at this point to be zero, when
0 = W1+W2
0 = kQ1Q3/x + k(-Q2)(Q3)/(x-r)
0 = kQ1Q3/x -kQ2Q3/(x-r)
-kQ1Q3/x = -kQ2Q3/(x-r)
Q1/x = Q2/(x-r)
Q1x-Q1r = Q2x
Q1x-Q2x = Q1r
(Q1-Q2)x = Q1r
x = Q1r/(Q1-Q2)
Note : Any one of the above cases is required.
