# A charge of 2 C is at the origin. How much energy would be applied to or released from a  -3 C charge if it is moved from  ( 6 , 6 )  to (-5 ,4 ) ?

Mar 27, 2018

We can calculate energy based on the formula
$U = \frac{1}{4 \pi {\epsilon}_{0}} \frac{{q}_{1} {q}_{2}}{r}$

The change in energy can be therefore calculated:
$\Delta U = {U}_{f} - {U}_{0} = \frac{1}{4 \pi {\epsilon}_{0}} \left(2 \setminus C\right) \cdot \left(- 3 \setminus C\right) \left(\frac{1}{r} _ f - \frac{1}{r} _ 0\right)$

The initial distance between the two points is ${r}_{0} = \sqrt{{6}^{2} + {6}^{2}} = 6 \sqrt{2}$
The final distance is
${r}_{f} = \sqrt{{5}^{2} + {4}^{2}} = \sqrt{41}$

I will assume the distance is in meters.

Therefore, the change in energy is
$\Delta U = - \frac{6 {C}^{2}}{4 \pi {\epsilon}_{0}} \left(\frac{1}{\sqrt{41} m} - \frac{1}{6 \sqrt{2} \setminus m}\right) \approx 2.07 \times {10}^{9} \setminus J$

This is a HUGE number, but to put it into context, a coulomb is an insane amount of charge and electromagnetism is really strong and these charges are really close to each other, so numbers this high aren't surprising.