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

Jan 23, 2016

$0.179 k \textcolor{w h i t e}{x} \text{units}$

#### Explanation:

The -3C charge moves from $P$ to $Q$.

I will work out the electric potential due to the charge at the origin at points $P$ and $Q$.

Then I will subtract these to get the potential difference between the 2 points in volts

If 1C of charge is moved through a potential difference of 1 Volt then 1 Joule of work is done.

So I will then multiply the potential difference by -3C to get the total work done.

Potential at P:

This is given by:

${E}_{P} = \frac{k . q}{d} _ 1$

$k$ is a constant.

To find ${d}_{1}$ we use Pyhtagoras:

${d}_{1} = \sqrt{{2}^{2} + {7}^{2}} = \sqrt{53}$

${d}_{1} = 7.28$

$\therefore {E}_{P} = \frac{k \times - 1}{7.28} = - \frac{k}{7.28} \text{ } \textcolor{red}{\left(1\right)}$

Potential at Q:

${E}_{Q} = k . \frac{q}{d} _ 2$

Again, we can use Pythagoras to find ${d}_{2} \Rightarrow$

${d}_{2} = \sqrt{{1}^{2} + {3}^{2}} = \sqrt{10}$

${d}_{2} = 3.16$

$\therefore {E}_{Q} = \frac{k \times - 1}{3.16} = \frac{- k}{3.16} \text{ } \textcolor{red}{\left(2\right)}$

To get the potential difference $\Delta E$ between the 2 points we subtract $\textcolor{red}{\left(1\right)}$ from $\textcolor{red}{\left(2\right)} \Rightarrow$

$\Delta E = - \frac{k}{7.28} - \left(- \frac{k}{3.16}\right)$

$\Delta E = k \left[\frac{1}{3.16} - \frac{1}{7.28}\right]$

$\Delta E = k \left[0.316 - 0.137\right] = 0.178 k$

To get the work done $W$ we multiply by the charge moved:

$W = - 3 \times 0.178 k = - 0.536 k \textcolor{w h i t e}{x} \text{units}$

The negative sign means work must be done on the charge to move it from $P$ to $Q$.

No units for distance are given. If it was in metres then $k = 9 \times {10}^{9} \text{m/F}$ and the answer would be in Joules.