# 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 ,-2 ) ?

Jan 12, 2016

Let ${q}_{1} = - 2 C$, ${q}_{2} = 4 C$, $P = \left(7 , 5\right)$, $Q = \left(3. - 2\right)$, and $O = \left(0.0\right)$
The distance formula for Cartesian coordinates is

d=sqrt((x_2-x_1)^2+(y_2-y_1)^2
Where ${x}_{1} , {y}_{1}$, and ${x}_{2} , {y}_{2} ,$ are the Cartesian coordinates of two points respectively.

Distance between origin and point P i.e $| O P |$ is given by.

$| O P | = \sqrt{{\left(7 - 0\right)}^{2} + {\left(5 - 0\right)}^{2}} = \sqrt{{7}^{2} + {5}^{2}} = \sqrt{49 + 25} = \sqrt{74}$

Distance between origin and point Q i.e $| O Q |$ is given by.

$| O Q | = \sqrt{{\left(3 - 0\right)}^{2} + {\left(- 2 - 0\right)}^{2}} = \sqrt{{\left(3\right)}^{2} + {\left(- 2\right)}^{2}} = \sqrt{9 + 4} = \sqrt{13}$

Distance between point P and point Q i.e $| P Q |$ is given by.

$| P Q | = \sqrt{{\left(3 - 7\right)}^{2} + {\left(- 2 - 5\right)}^{2}} = \sqrt{{\left(- 4\right)}^{2} + {\left(- 7\right)}^{2}} = \sqrt{16 + 49} = \sqrt{65}$

I'll work out the electric potential at points $P$ and $Q$.

Then I will use this to work out the potential difference between the two points.

This is the work done by moving a unit charge between the two points.

The work done in moving a $4 C$ charge between $P$ and $Q$ can therefore be found by multiplying the potential difference by $4$.

The electric potential due to a charge $q$ at a distance $r$ is given by:

$V = \frac{k \cdot q}{r}$

Where $k$ is a constant and its value is $9 \cdot {10}^{9} N {m}^{2} / {C}^{2}$.

So the potential at point $P$ due to charge ${q}_{1}$ is given by:

${V}_{P} = \frac{k \cdot {q}_{1}}{\sqrt{74}}$

The potential at $Q$ due to the charge ${q}_{1}$ is given by:

${V}_{Q} = \frac{k \cdot {q}_{1}}{\sqrt{13}}$

So the potential difference is given by:

${V}_{Q} - {V}_{P} = \frac{k \cdot {q}_{1}}{\sqrt{13}} - \frac{k \cdot {q}_{1}}{\sqrt{74}} = \left(k \cdot {q}_{1}\right) \left(\frac{1}{\sqrt{13}} - \frac{1}{\sqrt{74}}\right)$

So the work done in moving a ${q}_{2}$ charge between these 2 points is given by:

$W = {q}_{2} \left({V}_{Q} - {V}_{P}\right) = 4 \left(k \cdot {q}_{1}\right) \left(\frac{1}{\sqrt{13}} - \frac{1}{\sqrt{74}}\right) = 4 \left(9 \cdot {10}^{9} \cdot \left(- 2\right)\right) \left(\frac{1}{\sqrt{13}} - \frac{1}{\sqrt{74}}\right) = - 11.5993 \cdot {10}^{9}$

This is the work done on the charge.

There are no units of distance given. If this was in meters then the answer would be in Joules.