Question

A charge of `-2 C` is at the origin. How much energy would be applied to or released from a `4 C` charge if it is moved from `(7 ,5 )` to `(3 ,-9 )`?

Answer 1

See below.

Explanation:

The electrostatic potential energy between 2 charges follows from Coulomb's Law as:

`U = k (q_1q_2)/r` where k is Coulomb's constant `approx 8.9 times 10^9 mF^(-1)`

The distance between the charges changes from `sqrt(7^2 + 5^2) approx 8.6` to `sqrt(3^2 +(-9)^2) approx 9.48`.

So:

`Delta U = k q_1q_2 * (1/r_2 - 1/r_1)`

`= 8.9 times 10^9 (-2 times 4) * (1/sqrt(3^2 +(-9)^2) - 1/sqrt(7^2 + 5^2))`

`approx 7.7 times 10^8 J`

Given that a negative charge and a positive charge are pulled apart in this process, the potential energy of the system should increase, and energy needs to be applied to the system to achieve this.