# A charge of 5 C is at the origin. How much energy would be applied to or released from a  -2 C charge if it is moved from  ( 3 , 6 )  to ( -3 , 5) ?

Dec 19, 2016

The potential energy of the -2 C charge changes from $- 1.34 \times {10}^{10} J$ to $- 1.54 \times {10}^{10} J$, a change of $- 2.0 \times {10}^{9} J$ released.

#### Explanation:

Fortunately, by doing this problem using energy methods, we need only calculate the energy at the initial and final locations, and do not have to consider the (challenging) task of evaluating the force at each point along the path. (Calculus, anyone?)

At (3,6), the distance between the charges is $\sqrt{{3}^{2} + {6}^{2}}$ = $\sqrt{45}$

By Coulomb's law, the force is

$F = \frac{\left(9 \times {10}^{9}\right) \left(5\right) \left(2\right)}{45}$ = $2.0 \times {10}^{9}$ C (attractive)

and the potential energy is $P E = F \Delta d = \left(2.0 \times {10}^{9}\right) \times \sqrt{45}$= $- 1.34 \times {10}^{10} J$

At (3,-5) the distance between the charges is $\sqrt{{3}^{2} + {5}^{2}}$ = $\sqrt{34}$

By Coulomb's law, the force is

$F = \frac{\left(9 \times {10}^{9}\right) \left(5\right) \left(2\right)}{34}$ = $2.65 \times {10}^{9}$ C (attractive)

and the potential energy is $P E = F \Delta d = \left(2.65 \times {10}^{9}\right) \times \sqrt{34}$= $- 1.54 \times {10}^{10} J$

Thus, the change in energy has been $- 2.0 \times {10}^{9}$ J where I have affixed the negative as the PE is lower at (3,-5) than at (3,6) because the -2 C charge is now closer to the 5 C to which it is attracted.

(As an aside, in all the above, the location of zero potential energy is taken to be at $\infty$ for the negative charge. This is the usual convention. Thus, this charge has negative potential energy at all points a finite distance from the positive charge. A negative PE generally implies a bound state for the charges as energy must be supplied to separate them to large distance from each other.)