Parallel plate capacitor problem

  • Thread starter camel-man
  • Start date
  • #1
76
0

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.
 
Last edited by a moderator:

Answers and Replies

  • #2
gneill
Mentor
20,925
2,867
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).
 
  • #3
76
0
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.
 
  • #4
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,073
ε is always an abbreviation of the product of ε0 x εr,
the permittivities.
 
  • #5
gneill
Mentor
20,925
2,867
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?
 
  • #6
76
0
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 in to the C=εA/d when trying to solve for finding the seperation of 20.0nC
 
  • #7
gneill
Mentor
20,925
2,867
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 in to the C=εA/d when trying to solve for finding the seperation 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?
 
  • #9
gneill
Mentor
20,925
2,867
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?
 
  • #10
76
0
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
 
  • #11
gneill
Mentor
20,925
2,867
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.
 

Related Threads on Parallel plate capacitor problem

  • Last Post
Replies
8
Views
8K
  • Last Post
Replies
1
Views
7K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
7
Views
2K
Replies
2
Views
5K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
14
Views
8K
  • Last Post
Replies
3
Views
932
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
680
Top