# Homework Help: Voltage created by charged flat plates

1. Apr 1, 2012

### physickkksss

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

Two flat charged plates create a uniform electric field (E). What happens to voltage (V) when the distance (d) is doubled?

2. Relevant equations

Voltage in uniform electric field:
V = E*d

Capacitance equations:
V = Q/C
C = K * A/d

3. The attempt at a solution

The simple thing to do would be that since d is increased by 2, and V = Ed, then V goes up by 2.

But that does't make sense.....if you keep moving the plates apart would you keep getting an increase in voltage?

I think we need to incorporate the change in capacitance somehow....

Last edited: Apr 1, 2012
2. Apr 1, 2012

### tms

You are assuming that $\vec{E}$ stays the same if the plates are moved. Is that true?

3. Apr 1, 2012

### physickkksss

I'm not sure....

Doesn't E stay constant in a field with two flat charged plates?
I know if they were point charges than it would be E = K Q/r^2

Just it does not make sense that if d is doubled, then V should go up by 2 (or else you could keep moving plates apart to create more voltage).

It seems something else needs to be taken into account....either Capacitance, or E, or something...

4. Apr 1, 2012

### Staff: Mentor

There are certain assumptions concerning the dimensions of a capacitor that allow one to assume a uniform field between the plates. The resulting equations only hold so long as those assumptions are met. Investigate "capacitor edge effects".

5. Apr 1, 2012

### tms

And what happens if $r$ changes?

You're right, it doesn't make sense. And it's good you are checking your results for sanity. Why indeed should $V$ double when the distance doubles? Does it make any more sense for $V$ halve when the distance doubles? Review the meaning of $V$ and all should become clear.

Last edited: Apr 1, 2012
6. Apr 1, 2012

### Staff: Mentor

That is one way to generate a higher voltage. Charge some parallel plates, remove their link to the source, then increase the separation of the plates. It takes effort to move the plates apart; because being of opposite charge, there is attraction between the plates.

7. Apr 1, 2012

### physickkksss

So is that the simple answer then?....V increases by a factor of 2?

If the plates are very large, there are no changes in C or E?

8. Apr 1, 2012

### physickkksss

I noticed that there are two ways to go about it:

one way:
V = E*d.....in which case V increases by factor of 2

second way:

C = K A/d...which means C decreases by factor of 2

then V = Q/C....which means V increases by factor of 2

Both ways lead to the same answer (V increases by factor of 2). Is this a coincidence, or are these eqivalent ways to think about it?

9. Apr 1, 2012

### Staff: Mentor

On what basis do you say E remains fixed?

10. Apr 1, 2012

### physickkksss

From what I know E must be considered as one of two cases:

1) E between point charges ==> E = K Q/r^2

2) E between large flat charged plates ==> E remains constant

If two plates are moved backwards, would E change?

11. Apr 1, 2012

### rcgldr

In real life, with sufficient distance between the plates, the field approximates the field between two point charges. There's also a 3rd case, the field from a charged wire or between two charged wires. This 3rd case is rare, but the force related to the magnetic field between two wires carrying current is used to define what an ampere is.

Last edited: Apr 1, 2012
12. Apr 1, 2012

### physickkksss

hmmmm....I know things get more complicated in real life, but this is a typical "homework" type of question, where E is created from either 1) point charges, or 2) flat plates.

I am wondering whether:

V = E*d

and

C=K A/d
V=C/Q

are two equivalent ways to do this