A charge of -6 C is at the origin. How much energy would be applied to or released from a  9 C charge if it is moved from  (-2 ,3 )  to (2 ,-4 ) ?

Aug 10, 2017

The equation for Electrostatic potential energy between two charges is:

$E = {k}_{e} \frac{{Q}_{1} {Q}_{2}}{r}$

where ${k}_{e} \approx 8.99 \times {10}^{9} {\text{ JmC}}^{-} 2$

The change in energy by moving a charge is:

$\Delta E = {k}_{e} {Q}_{1} {Q}_{2} \left(\frac{1}{r} _ 2 - \frac{1}{r} _ 1\right)$

Compute ${r}_{1}$ using the point $\left(- 2 , 3\right)$

${r}_{1} = \sqrt{{\left(- 2\right)}^{2} + {3}^{2}}$

${r}_{1} = \sqrt{13} \text{m}$

Compute ${r}_{2}$ using the point $\left(2 , - 4\right)$

${r}_{2} = \sqrt{{2}^{2} + {\left(- 4\right)}^{2}}$

${r}_{2} = \sqrt{20} \text{m}$

We are given ${Q}_{1} = - 6 \text{C" and Q_2=9"C}$

DeltaE = (8.99xx10^9" JmC"^-2)(-6"C")(9"C")(1/(sqrt(20)"m")-1/(sqrt(13)"m"))

$\Delta E = 2.6 \times {10}^{10} J$