Question

A charge of `-1 C` is at the origin. How much energy would be applied to or released from a `-3 C` charge if it is moved from `( -1, -3 )` to `(-2 ,7 )`?

Answer 1

`0.179kcolor(white)(x)"units"`

Explanation:

The -3C charge moves from `P` to `Q`.

I will work out the electric potential due to the charge at the origin at points `P` and `Q`.

Then I will subtract these to get the potential difference between the 2 points in volts

If 1C of charge is moved through a potential difference of 1 Volt then 1 Joule of work is done.

So I will then multiply the potential difference by -3C to get the total work done.

Potential at P:

This is given by:

`E_P=(k.q)/d_1`

`k` is a constant.

To find `d_1` we use Pyhtagoras:

`d_1=sqrt(2^2+7^2)=sqrt(53)`

`d_1=7.28`

`:.E_P=(kxx-1)/7.28=-k/7.28" "color(red)((1))`

Potential at Q:

`E_Q=k.q/d_2`

Again, we can use Pythagoras to find `d_2rArr`

`d_2=sqrt(1^2+3^2)=sqrt(10)`

`d_2=3.16`

`:.E_Q=(kxx-1)/3.16=(-k)/3.16" "color(red)((2))`

To get the potential difference `DeltaE` between the 2 points we subtract `color(red)((1))` from `color(red)((2))rArr`

`DeltaE=-k/7.28-(-k/3.16)`

`DeltaE=k[1/3.16-1/7.28]`

`DeltaE=k[0.316-0.137]=0.178k`

To get the work done `W` we multiply by the charge moved:

`W=-3xx0.178k=-0.536kcolor(white)(x)"units"`

The negative sign means work must be done on the charge to move it from `P` to `Q`.

No units for distance are given. If it was in metres then `k=9xx10^9"m/F"` and the answer would be in Joules.