Question

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 `( 6 , 6 )` to `(-5 ,4 )`?

Answer 1

We can calculate energy based on the formula
`U = 1/(4pi epsilon_0) (q_1q_2) / r`

The change in energy can be therefore calculated:
`Delta U = U_f - U_0 = 1/(4pi epsilon_0) (2\ C) * (-3\ C) (1/r_f - 1/r_0)`

The initial distance between the two points is `r_0 = sqrt(6^2 + 6^2) = 6sqrt(2)`
The final distance is
`r_f = sqrt(5^2 + 4^2) = sqrt(41)`

I will assume the distance is in meters.

Therefore, the change in energy is
`Delta U = -(6C^2 )/ (4pi epsilon_0) (1 / (sqrt(41) m) - 1/(6sqrt2\ m) ) approx 2.07 times 10^9\ J`

This is a HUGE number, but to put it into context, a coulomb is an insane amount of charge and electromagnetism is really strong and these charges are really close to each other, so numbers this high aren't surprising.