# 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  (7 ,5 )  to (3 ,-9 ) ?

Mar 4, 2017

See below.

#### Explanation:

The electrostatic potential energy between 2 charges follows from Coulomb's Law as:

$U = k \frac{{q}_{1} {q}_{2}}{r}$ where k is Coulomb's constant $\approx 8.9 \times {10}^{9} m {F}^{- 1}$

The distance between the charges changes from $\sqrt{{7}^{2} + {5}^{2}} \approx 8.6$ to $\sqrt{{3}^{2} + {\left(- 9\right)}^{2}} \approx 9.48$.

So:

$\Delta U = k {q}_{1} {q}_{2} \cdot \left(\frac{1}{r} _ 2 - \frac{1}{r} _ 1\right)$

$= 8.9 \times {10}^{9} \left(- 2 \times 4\right) \cdot \left(\frac{1}{\sqrt{{3}^{2} + {\left(- 9\right)}^{2}}} - \frac{1}{\sqrt{{7}^{2} + {5}^{2}}}\right)$

$\approx 7.7 \times {10}^{8} J$

Given that a negative charge and a positive charge are pulled apart in this process, the potential energy of the system should increase, and energy needs to be applied to the system to achieve this.