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 ,-2 )`?

Answer 1

Let `q_1=-2C`, `q_2=4C`, `P=(7,5)`, `Q=(3.-2)`, and `O=(0.0)`
The distance formula for Cartesian coordinates is

`d=sqrt((x_2-x_1)^2+(y_2-y_1)^2`
Where `x_1, y_1`, and `x_2, y_2,` are the Cartesian coordinates of two points respectively.

Distance between origin and point P i.e `|OP|` is given by.

`|OP|=sqrt((7-0)^2+(5-0)^2)=sqrt(7^2+5^2)=sqrt(49+25)=sqrt74`

Distance between origin and point Q i.e `|OQ|` is given by.

`|OQ|=sqrt((3-0)^2+(-2-0)^2)=sqrt((3)^2+(-2)^2)=sqrt(9+4)=sqrt13`

Distance between point P and point Q i.e `|PQ|` is given by.

`|PQ|=sqrt((3-7)^2+(-2-5)^2)=sqrt((-4)^2+(-7)^2)=sqrt(16+49)=sqrt65`

I'll work out the electric potential at points `P` and `Q`.

Then I will use this to work out the potential difference between the two points.

This is the work done by moving a unit charge between the two points.

The work done in moving a `4C` charge between `P` and `Q` can therefore be found by multiplying the potential difference by `4`.

The electric potential due to a charge `q` at a distance `r` is given by:

`V=(k*q)/r`

Where `k` is a constant and its value is `9*10^9Nm^2/C^2`.

So the potential at point `P` due to charge `q_1` is given by:

`V_P=(k*q_1)/sqrt74`

The potential at `Q` due to the charge `q_1` is given by:

`V_Q=(k*q_1)/sqrt13`

So the potential difference is given by:

`V_Q-V_P=(k*q_1)/sqrt13-(k*q_1)/sqrt74=(k*q_1)(1/sqrt13-1/sqrt74)`

So the work done in moving a `q_2` charge between these 2 points is given by:

`W=q_2(V_Q-V_P)=4(k*q_1)(1/sqrt13-1/sqrt74)=4(9*10^9*(-2))(1/sqrt13-1/sqrt74)=-11.5993*10^9`

This is the work done on the charge.

There are no units of distance given. If this was in meters then the answer would be in Joules.