Question

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 `(7 , -4 )` to `(-3 , 5 )`?

Answer 1

`= -854 MJ`

Explanation:

There is a very similar question here
https://socratic.org/questions/a-charge-of-5-c-is-at-the-origin-how-much-energy-would-be-applied-to-or-released-4

A more expanded answer would strat with Coulomb's Law, so that the electric field `vec E` at any point in space due to the charge `Q_o` at the Origin is:

`vec E(vec r) = k Q_o/r^2 hat r`

This is spherically symmetrical, and:

• `hat r` is the unit radial vector `(vec r )/(abs ( vec r))`.

• `r = abs ( vec r)`, ie the distance of the point from the Origin

The change in potential [ie electrostatic potential energy per unit charge] in moving a unit positive test charge via any path from `|vec r_1| to oo` to `|vec r_2| = R` is the work done against the field along that path:

`Delta V = color(red)(-) int_(vec r_1)^(vec r_2) vec E(vec r) * d vec r`

The negative sign reflects that fact that the field is not doing the work, rather the work is being done against the field....

`= - k Q_o int_(vec r_1)^(vec r_2) 1/r^2 hat r * d vec r`

`= - k Q_o int_(vec r_1)^(vec r_2) 1/r^3 vec r * d vec r qquad triangle`

Now
`d (vec r * vec r) = d(r^2) = color(blue)(2r dr)`

But also true is
`d (vec r * vec r) = color(blue)(2 vec r * d vec r)` from the product rule and dot product commutivity

So we can write `triangle` as:

`= - k Q_o int_(r to oo)^(R) 1/r^2 d r`

`= k Q_o [ 1/r ]_(r to oo)^R`

`implies color(green)(Delta V_(oo to R) = V_R= (k Q_o) /R )`

....so `V_R` is the potential of a unit positive charge at position `vec r = R \ hat r` wrt the Origin.

At `(7, -4)`, `R = sqrt (7^2 + (-4)^2) = sqrt 65`

Likewise, at `(-3,5)`, we have `R = sqrt 34`

`Delta V = k Q_o ( 1/sqrt 34 - 1/sqrt 65 )`

The change in energy of a charge `q` is therefore
`Delta U = q Delta V = q \ k Q_o ( 1/sqrt 34 - 1/sqrt 65 )`

`approx (-1) 9.10^9 (2) ( 1/sqrt 34 - 1/sqrt 65 )`

`= -854 MJ`

Reality Check : Because the negatively charged particle has moved closer to the positively charged Origin, the particle has lost potential energy. Tick.