# A charge of 2 C is at the origin. How much energy would be applied to or released from a  -1 C charge if it is moved from  (7 , -4 )  to (-3 , 5 ) ?

Oct 3, 2016

$= - 854 M J$

#### Explanation:

There is a very similar question here
https://socratic.org/questions/a-charge-of-5-c-is-at-the-origin-how-much-energy-would-be-applied-to-or-released-4

A more expanded answer would strat with Coulomb's Law, so that the electric field $\vec{E}$ at any point in space due to the charge ${Q}_{o}$ at the Origin is:

$\vec{E} \left(\vec{r}\right) = k {Q}_{o} / {r}^{2} \hat{r}$

This is spherically symmetrical, and:

• $\hat{r}$ is the unit radial vector $\frac{\vec{r}}{\left\mid \vec{r} \right\mid}$.

• $r = \left\mid \vec{r} \right\mid$, ie the distance of the point from the Origin

The change in potential [ie electrostatic potential energy per unit charge] in moving a unit positive test charge via any path from $| {\vec{r}}_{1} | \to \infty$ to $| {\vec{r}}_{2} | = R$ is the work done against the field along that path:

$\Delta V = \textcolor{red}{-} {\int}_{{\vec{r}}_{1}}^{{\vec{r}}_{2}} \vec{E} \left(\vec{r}\right) \cdot d \vec{r}$

The negative sign reflects that fact that the field is not doing the work, rather the work is being done against the field....

$= - k {Q}_{o} {\int}_{{\vec{r}}_{1}}^{{\vec{r}}_{2}} \frac{1}{r} ^ 2 \hat{r} \cdot d \vec{r}$

$= - k {Q}_{o} {\int}_{{\vec{r}}_{1}}^{{\vec{r}}_{2}} \frac{1}{r} ^ 3 \vec{r} \cdot d \vec{r} q \quad \triangle$

Now
$d \left(\vec{r} \cdot \vec{r}\right) = d \left({r}^{2}\right) = \textcolor{b l u e}{2 r \mathrm{dr}}$

But also true is
$d \left(\vec{r} \cdot \vec{r}\right) = \textcolor{b l u e}{2 \vec{r} \cdot d \vec{r}}$ from the product rule and dot product commutivity

So we can write $\triangle$ as:

$= - k {Q}_{o} {\int}_{r \to \infty}^{R} \frac{1}{r} ^ 2 d r$

$= k {Q}_{o} {\left[\frac{1}{r}\right]}_{r \to \infty}^{R}$

$\implies \textcolor{g r e e n}{\Delta {V}_{\infty \to R} = {V}_{R} = \frac{k {Q}_{o}}{R}}$

....so ${V}_{R}$ is the potential of a unit positive charge at position $\vec{r} = R \setminus \hat{r}$ wrt the Origin.

At $\left(7 , - 4\right)$, $R = \sqrt{{7}^{2} + {\left(- 4\right)}^{2}} = \sqrt{65}$

Likewise, at $\left(- 3 , 5\right)$, we have $R = \sqrt{34}$

$\Delta V = k {Q}_{o} \left(\frac{1}{\sqrt{34}} - \frac{1}{\sqrt{65}}\right)$

The change in energy of a charge $q$ is therefore
$\Delta U = q \Delta V = q \setminus k {Q}_{o} \left(\frac{1}{\sqrt{34}} - \frac{1}{\sqrt{65}}\right)$

$\approx \left(- 1\right) {9.10}^{9} \left(2\right) \left(\frac{1}{\sqrt{34}} - \frac{1}{\sqrt{65}}\right)$

$= - 854 M J$

Reality Check : Because the negatively charged particle has moved closer to the positively charged Origin, the particle has lost potential energy. Tick.