# Work done is the area under a voltage against charge graph

1. Nov 24, 2013

### mrcotton

1. The problem statement, all variables and given/known data

2. Relevant equations

Area of a triangle is 0.5 × b × h
ΔW=VΔQ

3. The attempt at a solution

I drew a straight line graph from the origin with end point 4.5V 9.0 microF
energy is a work done = .5QV

There is a mark in the mark scheme for ΔW=VΔQ explained.

What do you think I need to say for this mark?

2. Nov 24, 2013

### Redbelly98

Staff Emeritus
Your physics textbook, and perhaps your class lecture notes, should have an explanation of electric potential and how it relates to charge as well as work or energy. I would take a look at that.

3. Nov 24, 2013

### mrcotton

Hi Redbelly thanks for responding.
I have just read in my notes that the energy stored in the capacitor is equal to the work done to force the extra charge onto the plates.

So ΔE=VΔQ
So V is changing and I guess if we make the increments very small in Q then the areas of each strip under the line add up to the total area under the triangle with base Q and Height V.
Thanks again