capacitors help

I was reading a physics book, see attached, and was wondering that when a capacitor discharges, the graph of current/charge/p.d. against time is an exponential decay graph. But when a capacitor charges up, how does the graph look like? Is it like the attached graph?

Thank you very much :smile:

Reply 1

Stonebridge

Original post by krisshP

IBut when a capacitor charges up, how does the graph look like? Is it like the attached graph?

Thank you very much :smile:

The graph of what? Charge? Pd? Current? All three?

Here's a question for you...

Two of them look like the one you have sketched, but one of them doesn't.

Which one doesn't?
Have a think.

Reply 2

krisshP

Original post by Stonebridge

The graph of what? Charge? Pd? Current? All three?

Here's a question for you...

Two of them look like the one you have sketched, but one of them doesn't.

Which one doesn't?
Have a think.

Not fully sure, but I'll try.

At first I imagine that there's no charge on the plate. So charge reaching the plates experiences little repulsion from other charges already on the plates as there's not much charge already on the plate. So charge builds up (like attached graph above) and so does the current in the circuit. Later on though, when there's a lot of charge on the plates, the incoming charge experiences high repulsion and so struggles to climb onto the plates. Hence rate of change of charge on the plates decreases as well as the current flowing in the circuit, like the graph attached above.

Not sure about p.d. though :confused:

, so I reckon p.d. is the one not following the above attached graph. Although I can't think of any logical explanation for this :confused:

Thanks

(edited 10 years ago)

Reply 3

Stonebridge

Original post by krisshP

This must be true. There is no charge initially on the capacitor and it builds up to some value Q where Q = CV

and so does the current in the circuit.

why?

Later on though, when there's a lot of charge on the plates, the incoming charge experiences high repulsion and so struggles to climb onto the plates. Hence rate of change of charge on the plates decreases as well as the current flowing in the circuit,

yes, current is rate of flow of charge onto the plates and this decreases as more charge gathers on the plates.

like the graph attached above.

Which graph? The one you drew shows an increase. Not a decrease.

Not sure about p.d. though :confused:

, so I reckon p.d. is the one not following the above attached graph. Although I can't think of any logical explanation for this :confused:

Thanks

Pd is given by V = Q/C so as charge builds up, so does the pd on the plates.

Reply 4

krisshP

Original post by Stonebridge

This must be true. There is no charge initially on the capacitor and it builds up to some value Q where Q = CV

why?

yes, current is rate of flow of charge onto the plates and this decreases as more charge gathers on the plates.

Which graph? The one you drew shows an increase. Not a decrease.

Pd is given by V = Q/C so as charge builds up, so does the pd on the plates.

Oh, I made a mistake for the current-time graph. The attached graph in the first post represents Q against t. Current is dQ/dt so is the gradient of the graph of Q against t. Initially there's little charge on the plates, so there's little repulsion to incoming charge, giving a high rate of flow of charge and so a high initial current. Later on though as the charge builds up on the plates, the repulsion to incoming charge builds up. Hence the rate of flow of charge to the plates decreases, like an exponential decay graph?

Oh it makes sense with p.d. now :smile:

Thanks