Why is the capacitance graph for V against T different to I against T?

Watch
Sayonara
Badges: 11
Rep:
?
#1
Report Thread starter 7 years ago
#1
Since Q = CV and C is constant, Q is proportional to V and so the graphs for V and Q (against T) should be the same shape.

Graph image:
http://www.antonine-education.co.uk/...tors/Cap_6.gif

However, the problem is...

Since V = IR, and R is constant, why isn't the graph for V against T the same shape as the graph for I against T for when charging a capacitor?

Why? This problem has hurt my head.

The current initially starts very high and drops down. So, it is different from the voltage and charge graphs, but why is the V = IR thing not working? After all, R is constant so V is proportional to I?

edit:
I know why the current against T graph looks like this and the logic behind it: http://i.gyazo.com/f9eb89b9aba6543139a0f945c9272ff8.png

However, what I don't know is why it looks like that when considering V = IR and that V is proportional to I?
0
reply
uberteknik
  • Study Helper
Badges: 21
Rep:
?
#2
Report 7 years ago
#2
(Original post by Sayonara)
Since Q = CV and C is constant, Q is proportional to V and so the graphs for V and Q (against T) should be the same shape.

Graph image:
http://www.antonine-education.co.uk/...tors/Cap_6.gif

However, the problem is...

Since V = IR, and R is constant, why isn't the graph for V against T the same shape as the graph for I against T for when charging a capacitor?

Why? This problem has hurt my head.

The current initially starts very high and drops down. So, it is different from the voltage and charge graphs, but why is the V = IR thing not working? After all, R is constant so V is proportional to I?
You cannot use Ohms law when dealing with capacitors. The capacitor plates are isolated and so have infinite resistance.

You need to think about this in a different way and about what's going on physically:

It helps if you have a picture of a capacitor in your head (two plates with a gap)
Like charges repel, unlike charges attract.
Current is a flow of electrons. I = Q/t
Think of the supply voltage as a 'pressure' forcing electrons around the circuit.

When the capacitor is uncharged and a voltage (pressure) is applied by the power supply across the plates, current starts to rush in to one plate because of the initial charge imbalance between the power supply and the uncharged capacitor plates. (the power supply is forcing electrons to move on to one of the plates and all of those electrons occupying the plate are pushing back against each other all the way back to the power source)

However, the electrons occupying the -ve plate cannot complete the circuit and flow directly to the supply negative terminal because the capacitor plates are insulated.

Instead however, as the -ve plate continues to fill with electrons (forced on to it by the power supply), they will set up a repulsion electric force across the plate-gap and in doing so, start to push electrons off the atoms of the other plate. It's these displaced electrons which flow back to the supply.

One plate thus gains a -ve charge and the other plate gains a +ve charge because the electrons are pushed away from that plate.


At the same time, the electrons filling up the -ve charge plate, also start to repel other electrons from being pushed onto that plate. i.e. a back-pressure is created by the electrons already occupying the -ve plate which (because like charges repel) ripples all the way back to the power source.

This back pressure is itself a voltage potential and it is this potential that rises as the capacitor continues to charge.

The supply voltage (pressure) remains constant, but the electric field building across the plate gap will reduce the current flowing into the capacitor because of the building back pressure. The more the charge on the capacitor, the less the charge current will be.

This is all saying that as Vc increases, then Ic decreases proportionally.

When the back pressure from the electric field stored between the capacitor plates is in equilibrium with the supply voltage pressure, no more charge can flow on to the capacitor. i.e. the charge current will be zero but the pd across the capacitor will then equal the supply voltage.
1
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Has your education been disrupted this academic year due to the pandemic?

Yes (2)
100%
No (0)
0%

Watched Threads

View All
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise