- #1
MattNotrick
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I have tried this set of questions which is from i-vii, I believe I am going wrong at a certain stage, as at question vi, the voltage is clearly wrong, below is the questions and my attempt. Sorry for the long post, any help on pointing out where I have gone wrong would be much appreciated, Thanks.
A parallel plate capacitor is constructed from two circular plates, radius 12cm and separated by a dieletric sheet with a thickness of 2 micro meters and relative permativity 6. The capacitor is charged to a potential difference of 110V
Questions
i) The area of one plate
ii) The capacitance
iii) The Charge on one plate
iv) The electric field between the plates
v) The energy stored in the capacitor
vi) The new potential difference if the dielectric is removed without changing the charge
vii) The plate separation is now changed to 1mm, find the energy compared to part v
Supplied within answers
i)Area = 0.0452m^2
ii)
c=εoεrA/d
C = (8.854x10^-12)(6)(0.0452) / (2x10^-6)
C=1.2x10^6F
iii)
Q=C/V
Q=(1.2x10^-6)(110)
Q=1.32x10^-4 C
iv)
E=v/d
E=110/2x10^-6
E=5.5x10^-5 J
v)
E=1/2CV^2
E=1/2(1.2x10^-6)(110)^2
E=7.26x10^-3 J
vi) (I feel I have gone wrong here, maybe earlier?)
I transpose C=Q/V to get V=C/Q to work out the new potential difference?
I need to work out the new capacitance with no dielectric so
C=(8.854x10^-12)(0.04522)/(2x10^-6)
C=2x10^-7 so
V=2x10^-7/1.32x10^-4
V= 1.51 x 10^-11
which surely cannot be right?
vii) (Need question vi correct to answer this one)
New capacitance = 2.4x10^-9
E=1/2CV^2
E=(2.4x10^-9)(Need answer from above)^2
E= Old Energy - New Energy
Thanks for anyone willling to help, Rick.
Homework Statement
A parallel plate capacitor is constructed from two circular plates, radius 12cm and separated by a dieletric sheet with a thickness of 2 micro meters and relative permativity 6. The capacitor is charged to a potential difference of 110V
Questions
i) The area of one plate
ii) The capacitance
iii) The Charge on one plate
iv) The electric field between the plates
v) The energy stored in the capacitor
vi) The new potential difference if the dielectric is removed without changing the charge
vii) The plate separation is now changed to 1mm, find the energy compared to part v
Homework Equations
Supplied within answers
The Attempt at a Solution
i)Area = 0.0452m^2
ii)
c=εoεrA/d
C = (8.854x10^-12)(6)(0.0452) / (2x10^-6)
C=1.2x10^6F
iii)
Q=C/V
Q=(1.2x10^-6)(110)
Q=1.32x10^-4 C
iv)
E=v/d
E=110/2x10^-6
E=5.5x10^-5 J
v)
E=1/2CV^2
E=1/2(1.2x10^-6)(110)^2
E=7.26x10^-3 J
vi) (I feel I have gone wrong here, maybe earlier?)
I transpose C=Q/V to get V=C/Q to work out the new potential difference?
I need to work out the new capacitance with no dielectric so
C=(8.854x10^-12)(0.04522)/(2x10^-6)
C=2x10^-7 so
V=2x10^-7/1.32x10^-4
V= 1.51 x 10^-11
which surely cannot be right?
vii) (Need question vi correct to answer this one)
New capacitance = 2.4x10^-9
E=1/2CV^2
E=(2.4x10^-9)(Need answer from above)^2
E= Old Energy - New Energy
Thanks for anyone willling to help, Rick.