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

`r_(B) = 3.34\ m`

Solution:

As per the question:

Point charge, q =
`5.45* 10^(- 8)\ C`

Test charge,
`q_(o) = 4.96* 10^(- 11)\ C`

Work done by the electric force,
`W_(AB) = - 9.05* 10^(- 9)\ J`

Now,

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

`V = (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)((1)/(r_(B) - (1)/(r_(A)))`

`- 9.05* 10^(- 9) = 9* 10^(9)* 5.45* 10^(- 8)* 4.96* 10^(- 11)((1)/(r_(B)) - (1)/(1.49)}`

`- 0.3719 = ((1)/(r_(B)) - 0.67}`

`r_(B) = (1)/(0.299) = 3.34\ m`