Potential difference between the parallel metal plates

[jamesjenson]

Homework Statement

Two metal plates are placed in parallel, 2mm apart. They are then connected to a fully charged 9V battery. Without discharging the plates, the battery is removed, after which the separation between them is increased to 1 cm. What is the new potential difference between the parallel metal plates?

Homework Equations

E=Vd E= Q/4ε∏r²

The Attempt at a Solution

I have no clue whatsoever, the prof never even went over this, just a hint in the right direction would be apperaited

 

[schaefera]

Ask yourself: with the battery removed, will the potential difference stay constant? Will the charge on each plate (and therefore electric field)?

There are several ways to find potential difference. If potential remains constant, use the one involving that. If electric field remains constant, use the one involving that instead.

 

[jamesjenson]

Ask yourself: with the battery removed, will the potential difference stay constant? Will the charge on each plate (and therefore electric field)?

There are several ways to find potential difference. If potential remains constant, use the one involving that. If electric field remains constant, use the one involving that instead.

Ok, so if there battery is removed then, and the plates don't touch anything, they would keep their charge, no? So the only change would be the distance between them?

 

[Fluxthroughme]

Ok, so if there battery is removed then, and the plates don't touch anything, they would keep their charge, no? So the only change would be the distance between them?

Yes, they keep their charge. It might also help to remember $E = \frac{\sigma}{\epsilon_0}$

 

[jamesjenson]

Yes, they keep their charge. It might also help to remember $E = \frac{\sigma}{\epsilon_0}$

ok, but doesn't {\sigma} mean Q/A ? and I don't have any info on the area of the plates here

 

[Fluxthroughme]

ok, but doesn't {\sigma} mean Q/A ? and I don't have any info on the area of the plates here

The point is that the area of the plates doesn't change. So if both Q and A stay the same, so does sigma, and we know $\epsilon_0$ is a constant.

 

[jamesjenson]

The point is that the area of the plates doesn't change. So if both Q and A stay the same, so does sigma, and we know $\epsilon_0$ is a constant.

ahhh! Ok, I don't see why I didnt see that, thanks!

 

[hms.tech]

Can we use the concept of capacitance here in this example :

C = eA/d
and
C = Q/V

I think that as "d" would be increased 5 times then "C" would decrease by five times.

As C decreases by 5 times then (since charge "Q" on the plates is conserved) "V" would increase by five times .

Thus (new) V = 45 V

Does that not satisfy all the electrostatics principles ?