Does Separating the Plates of a Capacitor Affect Its Charge?

[needhelpplease]

Homework Statement

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)

Homework Equations

C=Q/V
C=E0(Area/Distance)

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.

 

[DrClaude]

Homework Equations

C=Q/V
C=E0(Area/Distance)

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

 

[needhelpplease]

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

 

[needhelpplease]

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

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

 

[DrClaude]

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

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

 

[needhelpplease]

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

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

 

[DrClaude]

Area , hmm I looked at the main equation so i assumed its a variable

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

 

[needhelpplease]

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

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 I am just in a hurry) Studying for finals!

 

[DrClaude]

Now can we get on with the question please

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

 

[needhelpplease]

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

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

 

[DrClaude]

You got it!

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

 

[gneill]

Capacitance is decreased so charge is increased

I think you meant that charge is decreased, right?

 

[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.

 

[needhelpplease]

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.

Hmm the answer is decrease , I'm sorry i was typing too fast and i messed up. I got it , thank you! srry for not replying quickly but I only check this place if i got troubles I am not a regular visitor