Question

What is the electric potential from a charge of 3.0 x 10^-5 C at 15 cm and 65 cm? How much work must be done to carry a +2.3 x 10^-3 charge from 65 cm to 15 cm?

Answer 1

Potential `V` is:

`V = (k Q)/r`

• 15 cm

`V(0.15) = ((9 xx 10^(9)) (3xx10^(-5)))/(0.15) approx 1.8MV`

• 65 cm

`V(0.65) = ((9 xx 10^(9)) (3xx10^(-5)))/(0.65) approx 0.42MV`

The work required equals the change in potential energy `Delta U`, and `U = qV`

So:

`W = Delta U = q Delta V = 2.3xx10^(-3) cdot (1.8 - 0.42)cdot 10^6 = 3174 " J"`

Check the numbers yourself in yer calculator :)