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  ( -4, 3 )  to ( 1 , 4 ) ?

1 Answer
Apr 21, 2018

$W \approx 764790538 \text{J}$ or $764.8 \text{MJ}$

This energy is applied

Explanation:

First we need to find the distances from the origin to the 1C charge at both points.

${d}_{1} = \sqrt{{\left(- 4\right)}^{2} + {\left(3\right)}^{2}} = 5$

${d}_{2} = \sqrt{{\left(1\right)}^{2} + {\left(4\right)}^{2}} = \sqrt{17} \approx 4.123$

I'm assuming all distance measurements are in meters

The first observation you should make is that the 1C charge is being moved closer to the 2C charge, an action that, because of their repulsion, requires energy to be applied.

Let's start from our given: Coulomb's Law

$F = \frac{k {q}_{1} {q}_{2}}{d} ^ 2$

Where $k$ is Coulomb's constant, ${q}_{1}$ is the source charge at the origin, ${q}_{2}$ is the test charge, and $d$ is the distance between them.

From here, you can find the voltage (potential energy) at each point by substituting $V = {F}_{\text{electric}} d$ into Coulomb's Law.

$V = \frac{k {q}_{1} {q}_{2}}{d} ^ 2 \cdot d$

Simplifying

$V = \frac{k {q}_{1} {q}_{2}}{d}$

Where $V$ is voltage and NOT velocity (common mistake)

We also know that work is, in this problem, equal to the change in voltage (potential energy):

$\Delta V = W = \frac{k {q}_{1} {q}_{2}}{d} _ 2 - \frac{k {q}_{1} {q}_{2}}{d} _ 1$

Simplifying

$W = k {q}_{1} {q}_{2} \left(\frac{1}{d} _ 2 - \frac{1}{d} _ 1\right)$

From here, we can just put in the numbers:

$W = \left(\left(8.99 \cdot {10}^{9}\right) \cdot 2 \cdot 1\right) \left(\frac{1}{\sqrt{17}} - \frac{1}{5}\right)$

$W \approx 764790538 \text{J}$ or $764.8 \text{MJ}$

One Coulomb is a huge amount of charge; normally you would work problems with nothing larger than micro Coulombs.