Question

A positive point charge (q = +5.45 x 10-8 C) is surrounded by an equipotential surface A, which has a radius of rA = 1.49 m. A positive test charge (q0 = +4.96 x 10-11 C) moves from surface A to another equipotential surface B, which has a radius rB. The work done by the electric force as the test charge moves from surface A to surface B is WAB = -9.05 x 10-9 J. Find rB.

Answer 1

Answer:

$r_{B} = 3.34\ m$

Solution:

As per the question:

Point charge, q = $5.45\times 10^{- 8}\ C$

Test charge, $q_{o} = 4.96\times 10^{- 11}\ C$

Work done by the electric force, $W_{AB} = - 9.05\times 10^{- 9}\ J$

Now,

We know that the electric potential at a point is given by:

$V = \frac{kqq'}{r}$

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:

$W_{AB} = kqq_{o}(\frac{1}{r_{B} - \frac{1}{r_{A}}}$

$- 9.05\times 10^{- 9} = 9\times 10^{9}\times 5.45\times 10^{- 8}\times 4.96\times 10^{- 11}(\frac{1}{r_{B}} - \frac{1}{1.49}}$

$- 0.3719 = (\frac{1}{r_{B}} - 0.67}$

$r_{B} = \frac{1}{0.299} = 3.34\ m$