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Charge on Parallel Plate Capacitor

by mrlucky0
Tags: capacitor, charge, parallel, plate
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mrlucky0
#1
Sep20-07, 01:36 AM
P: 69
I am having a really tough time providing an argument for myself over why the answer is true:

The question:

A parallel plate capacitor is attached to a battery that supplies a constant potential difference. While the battery is still attached, the parallel plates are separated a little more.
Which statement describes what happens.

The solution:
ANSWER: Both the electric field and the charge on the plates decreases.

My reasoning:

The electric field decreasing, I can understand. I am imagining a small positive test charge inbetween the plates. As the plates are farther, the force on that test charge becomes weaker.

But what stumps me is why must the charges on the plates decrease? The problem states that the potential is kept constant, so what would influence the charges to decrease on the plates? If this can be reasoned using F=E*D, (F=force, E=electric field, D= distance) I'm certainly not understanding it.
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dynamicsolo
#2
Sep20-07, 01:45 AM
HW Helper
P: 1,663
Quote Quote by mrlucky0 View Post
But what stumps me is why must the charges on the plates decrease? The problem states that the potential is kept constant, so what would influence the charges to decrease on the plates? If this can be reasoned using F=E*D, (F=force, E=electric field, D= distance) I'm certainly not understanding it.

Here's one piece of it (I'm still cogitating about what is happening physically): you noted that the field strength decreases, since E = V/D, where V is the potential difference. [BTW, F=ED can't be right, both by the definitions of force and field strength and by the units. What *should* that be? (Don't change F or E.)] What *produces* the field between the plates? How is the field related to the charge on each plate?

(Now I [and *you*] have to think about what the charges are doing as the plates are separated further...)
learningphysics
#3
Sep20-07, 01:49 AM
HW Helper
P: 4,124
Voltage between the plates of a parallel plate capacitor = E*d where E is the electric field and d is the potential difference...

So as d increases E must decrease to maintain the same potential difference.

Also we know that Q = CV = (epsilon*A/d)*V

as d increases and V is kept constant, the right hand side decreases in magnitude. hence Q decreases.

mrlucky0
#4
Sep20-07, 02:04 AM
P: 69
Charge on Parallel Plate Capacitor

If this can be reasoned using F=E*D
Whoops. I actually meant V=E*D.

@LearningPhysics:

Thank you. Using Q=CV to justify makes sense to me now. That's what I was looking for.
dynamicsolo
#5
Sep20-07, 02:11 AM
HW Helper
P: 1,663
Quote Quote by mrlucky0 View Post
Whoops. I actually meant V=E*D.
The other equation that I was asking about was F = E*q .

As for the charge on the plates decreasing, you have probably also approximated a parallel-plate capacitor by two "infinite" conductive sheets of charge. The field of these sheets is related to their surface charge density. Since the plates don't change their surface area, and the decreasing field must be due to a decreasing surface charge density (since the field of the "infinite" sheets is *independent* of distance), that must mean that the charge on the two plates is decreasing.


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