Conductor inside capacitor check

[nissanztt90]

Homework Statement

An isolated conductor of thickness 'a' is placed between the plates of a parallel plate capacitor. Find the capacitance

Homework Equations

C = A/(4*pi*d)

The Attempt at a Solution

C1 = A/(4*pi*((d-a)/2))
C2 = A/(4*pi*((d-a)/2))

1/C = 1/C1 + 1/C2

Plugging in C1 and C2...

1/C = (8*pi*((d-a)/2)) / A

or

C = A / (8*pi*((d-a)/2)

If I am understanding this correctly...the conductor divides the capacitor into two capacitors in series...if someone can confirm this i appreciate it. TIA.

 

[berkeman]

Homework Statement

An isolated conductor of thickness 'a' is placed between the plates of a parallel plate capacitor. Find the capacitance

Homework Equations

C = A/(4*pi*d)

The Attempt at a Solution

C1 = A/(4*pi*((d-a)/2))
C2 = A/(4*pi*((d-a)/2))

1/C = 1/C1 + 1/C2

Plugging in C1 and C2...

1/C = (8*pi*((d-a)/2)) / A

or

C = A / (8*pi*((d-a)/2)

If I am understanding this correctly...the conductor divides the capacitor into two capacitors in series...if someone can confirm this i appreciate it. TIA.

Yes, it does divide it into two capacitors in series, each with a higher capacitance. If the dividing conductor is infinitely thin, then the change in capacitance should be what?

BTW, I'm not familiar with the capacitance equation you are using. Is it the form that is used in advanced physics units? The engineering form that I'm used to is:

C = \frac{\epsilon A}{d}

for a large thin capacitor (so edge fringe effects are negligible).

 

[nissanztt90]

The form i am using is just in CGS units...where the SI form is just divided by epsilon*4*pi I believe so that k = 1.

As the conductor thickness goes to 0, (d-a) will go to d, and the division by 2 will be negated by the factor of 8 as opposed to 4...so it will go back to the original equation.

 

[berkeman]

Thanks for the note about the cgs approach -- I was thinking that was the difference. And you are correct on the limit -- makes sense after all. It's a good way to check your equations, but it looks like you already knew that.

 

[nissanztt90]

Yes i did...thanks again for the check.