A charge of 5 C is at the origin. How much energy would be applied to or released from a  -4 C charge if it is moved from  ( 8, 2 )  to (9 , 6 ) ?

Dec 30, 2015

$0.43 k$ units of work applied to the charge.

Explanation:

http://tube.geogebra.org/student/m2354629

We can work out the electric field potential at point R.

Then we work out the electric field potential at point Q.

We can then get the work done in moving a charge of $- 4 C$ between these 2 points by multiplying the charge by the potential difference.

Potential at point R:

${V}_{R} = - \frac{k .5}{r} \text{ } \textcolor{red}{\left(1\right)}$

From the diagram you can see that:

${r}^{2} = {8}^{2} + {2}^{2} = 68$

$\therefore r = \sqrt{68} = 8.24$

So from $\textcolor{red}{\left(1\right)}$:

${V}_{R} = - k \frac{.5}{8.24} = - k .0 .607$

Potential at point Q:

${V}_{Q} = - \frac{k .5}{d} \text{ } \textcolor{red}{\left(2\right)}$

From the diagram you can see that:

${d}^{2} = {8}^{2} + {6}^{2} = 100$

$\therefore d = \sqrt{100} = 10$

So from $\textcolor{red}{\left(2\right)}$:

${V}_{Q} = - \frac{k .5}{10} = - \frac{k}{2}$

So the potential difference $E$ between R and Q $\Rightarrow$

$E = - k 0.607 - \left(- k 0.5\right)$

$E = - 0.107 k$

The work done in bringing a charge of $- 4 C$ through this potential difference is given by:

$W = - 0.107 k \times \left(- 4\right)$

$\therefore W = 0.43 k$ units

This means work would have to be applied to the charge.

There are no units of distance given. If you were using metres then $k = 9 \times {10}^{9} \text{m/F}$ and the answer would be in Joules.