# Capacitance and Capacitors

Tags:
1. Apr 12, 2017

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

The plates of a parallel-plate capacitor are maintained with constant voltage by a battery as they are pulled apart. During this process, the amount of charge must.... (decrease , increase , remains the same)
2. Relevant equations
C=Q/V
C=E0(Area/Distance)

3. The attempt at a solution
I know the answer but i don't know how to do it.
Basically if area is increased , capacitance increases because they are directly proportional. if capacitance increases then charge increases also because they are directly proportional.

The answer is decrease by the way.

2. Apr 12, 2017

### Staff: Mentor

You have four parameters describing the capacitor. In the scenario of the problem, which stay constant and which can vary?

3. Apr 12, 2017

Voltage is constant. while capacitance , area , charge are variables

4. Apr 12, 2017

Voltage is constant. while capacitance , area , charge are variables

5. Apr 12, 2017

### Staff: Mentor

What about distance? And please explain what you think the area is.

6. Apr 12, 2017

Right , distance is the main variable because it changes. Area , hmm I looked at the main equation so i assumed its a variable

7. Apr 12, 2017

### Staff: Mentor

But do you understand what it represents in an actual plate capacitor?

8. Apr 12, 2017

Yeah , it's the surface area of the plate. Now can we get on with the question please , i have like 50 more to solve. (im not being rude im just in a hurry) Studying for finals!

9. Apr 12, 2017

### Staff: Mentor

You should have all you need to figure out the answer now.

10. Apr 12, 2017

OH CRAP. I messed up. Area has nothing to do with it. Distance is increased , Capacitance is decreased so charge is increased. Thank man! Really appreciate it

11. Apr 12, 2017

### Staff: Mentor

You got it!

Edit: thanks to @gneill for reading more carefully than me.

Last edited: Apr 12, 2017
12. Apr 12, 2017

### Staff: Mentor

I think you meant that charge is decreased, right?

13. Apr 12, 2017

### epenguin

If the plates of the capacitor were 25 m apart you wouldn't expect it to be a good capacitor. But as you approached the plates it would get better. Why? Because the work the battery would have to do separating the charges is decreased by their being less separated until with a good capacitor they are little separated at all but are close to each other, a short distance facing through the dielectric.

So you should think physically not just formulaicially (although maybe you do as you said you knew the answer). You'd do well to do it more thoroughly than I have, calculate some forces with typical charges, numbers and molar amounts of electrons etc.

14. Apr 23, 2017