# Homework Help: Calculating an expression for the charge in a capacitor?

1. Aug 15, 2010

### Blimeyson

1. The problem statement, all variables and given/known data
hi i'm new and thanks in advance!

I'm trying to work out a problem i've been given but having difficulties! I am given a scope for the current in a capacitor and i'm trying to determine the expression of charge in the capacitor firstly, then when t = 10ms

2. Relevant equations

the Applitude is 6 and the scope rises from 0 to 6 in 5ms then stays at this value past the 20ms given.

any help is very much appresiated!

3. The attempt at a solution

so far i have the gradient = 6 (applitude) / 5mS = 1200 so i expect di/dt = 1200

and i know the expression for charge is Q= 1/C integral( i dt)

Last edited: Aug 15, 2010
2. Aug 15, 2010

### xcvxcvvc

can you post the problem how it was originally worded? what you're posting makes no sense

3. Aug 15, 2010

### Blimeyson

Sorry it is;

The current through a capacitor was displayed on a scope as shown below:

The Scope is a graph with current against time. The wave starts at 0 increases to 6 amps after 5 milli seconds then stays at this level up to the final 20 milli seconds.

Then the question goes on to say "choose an appropriate method to determine the expression for charge in the capacitor and so determine the charge in the capacitor when t=10 milli seconds. Hope this helps?

4. Aug 15, 2010

### xcvxcvvc

Does it increase to 6 amps linearly? Does the capacitor have zero charge initially?

5. Aug 15, 2010

### Blimeyson

the graph does not show charge it just has current I against time on the x axis. yes when time is zero current is also zero, it is linear to 5milli seconds, at a peak current of 6 amps then it stas at this level for a further 15 milli seconds. sorry it doesnt give anything away about the capacitor.

Quite a hard one!

6. Aug 15, 2010

### stevenb

Not so hard, but it's always a matter of looking at it from the right point of view.

What is the definition of current? Current is dQ/dt, or the time rate of change of charge. This means that charge can be found by integrating the current (i.e. adding up the area under the current curve). When you integrate, there is always an arbitrary integration constant to consider. This represents the initial charge. So charge on the capacitor is its initial charge plus an integration of current (or area under the current curve).

In the question, it's not clear what they mean by "choose an appropriate method". The method could be the integral equation you write down, or it could be more like a prescription to estimate the area under the curve graphically, which is not hard to do given the simple waveform.

7. Aug 16, 2010

### xcvxcvvc

Without knowing the initial charge on the capacitor, this problem is impossible. Tell your teacher to go back to school, and laugh in his face. Or you could solve it with the most probable case that the initial charge equals zero. As the post above me detailed, it is the integration of current -- or from calculus II you should remember it is the area under the current graph.

So if we want to know the charge on the capacitor at time T the answer is

(area under current curve from -infinity to T) + (initial current)

which, with the assumption that the initial charge is zero and that there is no relevant missing current waveform missing prior to t = 0, becomes

(area under current curve from 0 to T)

8. Aug 16, 2010

### stevenb

I'm sure you don't mean that literally, but I can't help but point out that it's not a good idea to insult a teacher this way, and it's not clear that the teacher is not aware of such a basic fact. The teacher may be asking the student to think a little, or there may be some additional information that was not given to us.

The crux of the question is to find the method. The method can include a way to either estimate the initial charge, or there may be a way to know that the initial charge is zero. If these approaches are not possible, then the answer can be be expressed with a Qo which represents the initial charge which is unknown now, but can be found after the fact if the final charge is left on the capacitor.

Even if the teacher made a bonehead mistake (which anyone can do by the way, and rarely requires going back to school to remedy, - usually a self-administered head-slap fixes the problem), the student will get much more credit for answering the question as intelligently as possible and expressing any issue with the phasing of the question.