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 `( -4, 0 )` to `( -6 ,-3 )`?

Answer 1

change in U = dU = `k Q q(1/(r_text(initial)-r_text(final)))`
`dU ~= -1.99 times 10^10 J`
Therefore around 19.9 GJ has been released by moving the charge.

Explanation:

Electrical potential energy = `U = (k Q q) / r`
k = Coulomb's constant `~= 8.9876 times 10^9 J m C^-2`
`r_text(initial) = sqrt((-4)^2+0^2)m=sqrt(16)m=4m`
`r_text(final)=sqrt((-6)^2+(-3)^2)m=sqrt(45)m=3sqrt(5)m`

change in U = dU = `k Q q(1/(r_text(initial)-r_text(final)))`

`dU = (8.9876 times 10^9 J m C^-2)(2C)(3C)(1/(4m-3sqrt(5)m))`

`dU ~= (5.3923 times 10^9 J m) (1/(4m-3sqrt(5)m))`

`dU ~= -1.99 times 10^10 J`

Therefore around 19.9 GJ has been released by moving the charge.

Answer 2

proceed as follows

Explanation:

Let the 2C charge is at origin O (0,0)
The initial position of 3C is at A (-4,0)
The final position of 3C is at B (--6,-3)
find distance OA,OB
find initial Pot Energy of the system when 3C is at A
then when at B using formula `(q_1*q_2)/(4piepsilonr`
calculate the difference ----Final - initial