Parallel plate capacitor problem

[camel-man]

Homework Statement

A parallel plate capacitor is constructed with plates of area .0280 m^2 and separation .550mm

Find the charge on each plate when potential difference between plates is 20.1V.

what separation would be necessary to give a charge of 20 nC on each plate?

Homework Equations

Im not sure that is where i am confused[/b]

The Attempt at a Solution

I do not even know which equations to use... I think it might be C=εA/d
and another question is what is the difference between C=Q/V and C=εA/d? When should i choose between using the two.?
but I'm taking a shot in the dark.

 

[gneill]

Hint: Use both equations for this problem.

The first equation relates the physical geometry of the device to its electrical property (capacitance).

The second equation relates the electrical quantities of voltage and charge to the given capacitance (regardless of geometry).

 

[camel-man]

My thoughts are this.. Use C=εA/d where ε=Kε0 and K=.5mv^2 (what is mass in this case?)

so we have multiple unknowns to get to a correct answer. If i can solve for K then I could get C and then plug it into Q=CV to get charge.

 

[NascentOxygen]

ε is always an abbreviation of the product of ε₀ x ε_r,
the permittivities.

 

[gneill]

My thoughts are this.. Use C=εA/d where ε=Kε0 and K=.5mv^2 (what is mass in this case?)

Ah, no. Here ε is a physical constant, the permittivity of free space. It's not related to any kinetic energy or mass or velocity in this problem. I think you may be confusing the constant K from Coulomb's law with the variable name KE which represent kinetic energy.

so we have multiple unknowns to get to a correct answer. If i can solve for K then I could get C and then plug it into Q=CV to get charge.

There is no K. Start with what you've been given. You are given the physical geometry of the capacitor and you have an equation that related that geometry to the capacitance. Why not find the capacitance?

 

[camel-man]

Okay, so I solved for C=εA/d I got 1.26x10^-11

I plugged C into Q=CV and got 2.53x10^-10

does this look right so far?

If it is, than I suppose I will factor that into the C=εA/d when trying to solve for finding the separation of 20.0nC

 

[gneill]

Okay, so I solved for C=εA/d I got 1.26x10^-11

I plugged C into Q=CV and got 2.53x10^-10

does this look right so far?

If it is, than I suppose I will factor that into the C=εA/d when trying to solve for finding the separation of 20.0nC

Your value for capacitance looks odd to me. Did the original problem state the plate area to be 0.0280 square meters, or 0.0280 meters square?

 

[camel-man]

Meters square

 

[gneill]

Okay. That makes a difference! Your calculations look okay so far. Be sure to include units on your numerical results.

For the next part you are given both charge and voltage and you need to find the required plate separation. How might you do that with your given equations?

 

[camel-man]

I got this d = 6.97x10^-9

This doesn't seem right to me. This is how I came to that answer...

d = εAV/Q

and I used 20.0x10^-6 C for the Q at the bottom

 

[gneill]

I got this d = 6.97x10^-9

This doesn't seem right to me. This is how I came to that answer...

d = εAV/Q

and I used 20.0x10^-6 C for the Q at the bottom

Looks like a unit conversion issue (power of ten). The digits are okay.