# 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 , 4 )  to (-3 ,2 ) ?

Aug 5, 2016

The key to this problem is to find the change in the distance between the two charges. The change in energy will be given by $U = \frac{k {q}_{1} {q}_{2}}{r}$. The energy lost to the environment is $1.5 \times {10}^{10}$ $J$.

#### Explanation:

Initial separation:

${r}_{1} = \sqrt{{\left(6 - 0\right)}^{2} + {\left(4 - 0\right)}^{2}} = \sqrt{52} \approx 7.2$ $m$ (we have to assume the units are $m$)

Final separation:

${r}_{2} = \sqrt{{\left(- 3 - 0\right)}^{2} + {\left(2 - 0\right)}^{2}} = \sqrt{13} \approx 3.6$ $m$

Difference: $7.2 - 3.6 = 3.6$ $m$

The change in the energy will be:

$\Delta U = \frac{k {q}_{1} {q}_{2}}{\Delta r} = \frac{9 \times {10}^{9} \times 2 \times - 3}{3.6} = - 1.5 \times {10}^{10}$ $J$.

Since the charges are opposite and they have become closer, the energy of the charge system has decreased, so energy will have been lost to the environment.