A charge of 2 C is at the origin. How much energy would be applied to or released from a  -3 C charge if it is moved from  (3 ,7)  to (8,2 ) ?

Nov 14, 2017

The energy required to move the $- 3$ $C$ charge against the attractive interaction of $+ 2$ $C$ charge at origin, from the coordinate point $\left(3 m , 7 m\right)$ to the coordinate point $\left(8 m , 2 m\right)$ is:
$E = \setminus \Delta U = 541.44 \setminus \times {10}^{6}$ $J = 541.44$ $M J$

Explanation:

The electric potential of a point charge '$Q$' at a distance '$r$' from it is given by:
V(r)= kQ/r; \qquad k=1/(4\pi\epsilon_0) = 8.99\times10^9 (Nm^2)/C^2

When a test charge ${q}_{0}$ is held at this point, its potential energy at this position is given as: $U \left(r\right) = {q}_{0} V \left(r\right)$.

Given: $Q = 2$ C; \qquad q_0 = -3 $C$.

Since the Electric Field is a conservative force field, the change in potential energy as you go from one point to another depends only on the end points and is independent of the path you choose to go.

So calculate the potential energies at the end points and find the difference.

When you take the point charge (${q}_{0}$) from a starting point to an end point at distances ${r}_{i}$ and ${r}_{f}$, respectively, from the source charge $Q$, change in its electric potential is :

$\setminus \Delta V = {V}_{f} - {V}_{i} = k \frac{Q}{r} _ f - k \frac{Q}{r} _ i = k Q \left(\frac{1}{r} _ f - \frac{1}{r} _ i\right)$ ...... (1)

Change in electric potential energy is:
$\setminus \Delta U = {U}_{f} - {U}_{i} = {q}_{0} \left({V}_{f} - {V}_{i}\right) = k Q {q}_{0} \left(\frac{1}{r} _ f - \frac{1}{r} _ i\right)$ ...... (2)

Distance to Start & End points : The starting point coordinates and the end point coordinates of the test charge are $\left(3 m , 7 m\right)$ and $\left(8 m , 2 m\right)$ respectively. So their distances from the source charge are:
${r}_{i} = \setminus \sqrt{{\left(3 m\right)}^{2} + {\left(7 m\right)}^{2}} = 7.6158$ $m$.
${r}_{f} = \setminus \sqrt{{\left(8 m\right)}^{2} + {\left(2\right)}^{2}} = 8.2462$ $m$.

Substituting the values of $Q , {q}_{0} , {r}_{i}$ and ${r}_{f}$ in (2) we get:

$\setminus \Delta U = 541.44 \setminus \times {10}^{6}$ $J = 541.44$ $M J$

So the potential energy increase by $541.44$ $M J$. This must be the energy required to do this movement.