Charge on Parallel Plate Capacitor

In summary, when a parallel plate capacitor is attached to a battery that supplies a constant potential difference and the plates are separated further, both the electric field and the charge on the plates decrease. This can be reasoned using the equation V=E*d, where V is the potential difference, E is the electric field, and d is the distance between the plates. The potential difference remains constant, so as d increases, E must decrease. Additionally, the charge on the plates can be justified using the equation Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference. As d increases and V is kept constant, the right hand side of the equation decreases, resulting in a decrease in charge. This decrease
  • #1
mrlucky0
69
1
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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  • #2
mrlucky0 said:
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...)
 
  • #3
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.
 
  • #4
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.
 
  • #5
mrlucky0 said:
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.
 

1. What is a charge on a parallel plate capacitor?

A charge on a parallel plate capacitor refers to the amount of electric charge that is stored on the capacitor's plates. This charge is created by the separation of positive and negative charges on the plates, creating an electric field between them.

2. How is the charge on a parallel plate capacitor calculated?

The charge on a parallel plate capacitor can be calculated by multiplying the capacitance of the capacitor by the voltage across it. This can be expressed as Q = CV, where Q is the charge, C is the capacitance, and V is the voltage.

3. What factors affect the charge on a parallel plate capacitor?

The charge on a parallel plate capacitor is affected by the distance between the plates, the surface area of the plates, and the type of material between the plates. It is also affected by the voltage applied across the capacitor.

4. How does the charge on a parallel plate capacitor change with a changing voltage?

As the voltage across a parallel plate capacitor changes, the charge on the capacitor also changes. This is because the capacitance of the capacitor is constant, so when the voltage increases, the charge also increases and vice versa.

5. Why is the charge on a parallel plate capacitor important?

The charge on a parallel plate capacitor is important because it is what allows the capacitor to store electrical energy. This stored charge can then be released to power electrical devices or circuits, making capacitors an essential component in many electronic devices.

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