# A charge of -2 C is at the origin. How much energy would be applied to or released from a  4 C charge if it is moved from  (-5 ,1 )  to (3 ,-9 ) ?

Initial distance between the $- 2 C$ charge and $4 C$ charge is $\sqrt{{\left(- 5\right)}^{2} + {\left(1\right)}^{2}} m$ or, $5.099 m$
And,final distance is $\sqrt{{\left(3\right)}^{2} + {\left(- 9\right)}^{2}} m$ i.e $9.486 m$
That means,distance has been increased in this process,against an attractive force between a negative charge and a positive charge,so energy has to be applied for moving against the attractive force.,and the value is $- U 2 - \left(- U 1\right)$ (where, $- U 2$ is the final energy and $- U 1$ is the initial energy of the charge)
$9 \cdot {10}^{9} \cdot \left(- 2\right) \cdot 4 \cdot \left(\frac{1}{9.486} - \frac{1}{5.099}\right) J$ or, $6.552 \cdot {10}^{9} J$