# 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?

Apr 28, 2017

Potential $V$ is:

$V = \frac{k Q}{r}$

• 15 cm

$V \left(0.15\right) = \frac{\left(9 \times {10}^{9}\right) \left(3 \times {10}^{- 5}\right)}{0.15} \approx 1.8 M V$

• 65 cm

$V \left(0.65\right) = \frac{\left(9 \times {10}^{9}\right) \left(3 \times {10}^{- 5}\right)}{0.65} \approx 0.42 M V$

The work required equals the change in potential energy $\Delta U$, and $U = q V$

So:

$W = \Delta U = q \Delta V = 2.3 \times {10}^{- 3} \cdot \left(1.8 - 0.42\right) \cdot {10}^{6} = 3174 \text{ J}$

Check the numbers yourself in yer calculator :)