Answer:
[tex]r_{B} = 3.34\ m[/tex]
Solution:
As per the question:
Point charge, q = [tex]5.45\times 10^{- 8}\ C[/tex]
Test charge, [tex]q_{o} = 4.96\times 10^{- 11}\ C[/tex]
Work done by the electric force, [tex]W_{AB} = - 9.05\times 10^{- 9}\ J[/tex]
Now,
We know that the electric potential at a point is given by:
[tex]V = \frac{kqq'}{r}[/tex]
where
r = separation distance between the charges.
Also,
The work done by the electric force i moving a test charge from point A to B in an electric field:
[tex]W_{AB} = kqq_{o}(\frac{1}{r_{B} - \frac{1}{r_{A}}}[/tex]
[tex]- 9.05\times 10^{- 9} = 9\times 10^{9}\times 5.45\times 10^{- 8}\times 4.96\times 10^{- 11}(\frac{1}{r_{B}} - \frac{1}{1.49}}[/tex]
[tex]- 0.3719 = (\frac{1}{r_{B}} - 0.67}[/tex]
[tex]r_{B} = \frac{1}{0.299} = 3.34\ m[/tex]
