Conceptual Problem about Capacitance

[sj005]

I think I know the answers to the question I am going to ask but I am just going to make sure. This is going to be quite simple for most of the experts here so here goes:

If we have a parallel plate capacitor and we double the distance between the plates while keeping the charge Q on the plates fixed. What will happen to:

a)The Capacitance
b)The Potential Difference
c)The Electric Field between the Plates
d)The Stored Energy

I studied the formulas/concepts and the answers I predicted were

a) The capacitance will be halved
b) The potential difference will double
c) The electric field will double
d) The energy stored will be twice as much

Are my answers correct or are they completely off?

Thanks in advance :)

 

[tiny-tim]

welcome to pf!

hi sj005! welcome to pf!

Are my answers correct …

maybe!

give reasons (ie a quick equation in each case)

 

[ZealScience]

Welcome to world of physics!

I think you are correct. Using derived equations C=Aε0/d and U=1/2CV^2.

 

[sj005]

Thanks for the welcome message guys :). I've heard a lot about Physics Forums-everyone tells me its awesome :D. In reply to tiny-tim, I derived my answers from these formulas:

a)C=Aε0/d - Capacitance is based on the geometry. The area of the plates didn't change, and the permitivity of free space ε0 is a constant so if the distance is doubled, the capacitance would become half its initial value.
b)Q=CV - The capacitance was halved(part a) so if the charge Q is to remain the same(as stated in the problem), the voltage V must become twice its original value.
c)This one seems most confusing to me and I just considered that the electric field varies according to the voltage. I don't think I got this one right.
d) E(stored energy)=(1/2)(C)(V^2) so if the capacitance is halved and the voltage doubles, we could write it as 1/2(C/2)(2V)^2 which would become
=(C/4)(2^2)(V^2)
=(C/4)(4)(V^2)
The 4 cancels in the numerator and denominator leaving:
=(C)(V^2) which is twice the original stored energy

Is my way on thinking correct or should I consider using more physical reasoning or perhaps different formulas?

 

[Gordianus]

To sj005:

How do you account for the extra energy stored in the capacitor?

 

[tiny-tim]

hi sj005!

(try using the X² icon just above the Reply box )

a) fine (except more generally you can say C=Aε/d, for a general dielectric)

b) fine, though you can also get V directly as ∫ E dx

c) no, E = D/ε, which depends only on Q and A … see your book, or the PF Library on capacitor

d) fine, but learn the other formulas such as 1/2 QV

 

[sj005]

Thanks again tiny tim :) . I guess I do need to go back to my book and study the concepts a little more. :D

 

[ZealScience]

hi sj005!

(try using the X² icon just above the Reply box )

a) fine (except more generally you can say C=Aε/d, for a general dielectric)

b) fine, though you can also get V directly as ∫ E dx

c) no, E = D/ε, which depends only on Q and A … see your book, or the PF Library on capacitor

d) fine, but learn the other formulas such as 1/2 QV

Sorry, I didn't read the third one carefully. In capacitors, electric field is uniform, so E=V/d. As V and d both doubled (as you sad yourself), E is constant, no need for other formulae.

The last, equation actually derives E=1/2CV^2 via substituting Q=CV