# Simple and pedantic question about a problem involving q=it

1. Sep 1, 2015

### s3a

1. The problem statement, all variables and given/known data
The problem and solution are attached as TheProblemAndSolution.jpeg.

2. Relevant equations
q = i × t

3. The attempt at a solution
For part (b), I just wanted to ask:
What does q(o-) = 0 (which is at the bottom of the solution) mean? Is the author trying to say $q(0^{-}) = 0$ in the sense that he or she is trying to say that he or she is assuming that q = 0 for any t < 0?

Any input would be greatly appreciated!

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2. Sep 1, 2015

### RUber

I think it means, more generally, that q=0 immediately before t=0. Thus it is realistic that at t=0, q=0.
What would happen if q>0 immediately preceding the time interval? You might need something like q(t) = q_0 + it.

3. Sep 1, 2015

### s3a

Thanks for the response.

Why did the author focus only on the immediate moment before t = 0 instead of any moment before t = 0, though?

Is there some kind of significance to ignoring values less than t = 0 that aren't the immediate moment before t = 0?

Last edited: Sep 1, 2015
4. Sep 1, 2015

### RUber

I think it may have to do with some continuity requirement.

5. Sep 7, 2015

### s3a

Even if you're not 100% sure, could you please be more specific as to what you're thinking? Perhaps what you're thinking could be just what I need to get the epiphany I'm looking for.

6. Sep 8, 2015

### RUber

As with most things, q = i*t is likely a simplification of something like q(t) = q_0+ i*t. The i*t piece is clearly 0 at t=0, and most physical systems require continuity. To that end, it doesn't matter what the system was doing in negative time, as long as it was zeroed out before this time interval started. If there were a charge present at t=0-, then some of that charge would be carried into the time interval we are interested in.

7. Sep 10, 2015

### s3a

So, that assumption is just used to ignore and not have to use a y-intercept, in order to simplify the equation?

8. Sep 10, 2015

### RUber

Yes.

9. Sep 10, 2015

### s3a

Okay, I get it now. :)

Thanks!