Parallel Plate Capacitor Question (Many Attempts)

[MattNotrick]

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.

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.

 

[gneill]

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

Okay up to here, but you missed the sign of the exponent in the capacitance above.

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

Watch your units. You divided volts (Joules/Coul) by distance (meters). Probably better to just leave the units as V/m.

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?)

Pretty much okay so far.

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

You could simply have 'reversed' the effect of the relative permittivity of the dielectric by dividing the previous capacitance by it: $$C/\epsilon$$

V=2x10^-7/1.32x10^-4
V= 1.51 x 10^-11
which surely cannot be right?

You want V = Q/C, but you performed V = C/Q.

vii) (Need question vi correct to answer this one)

New capacitance = 2.4x10^-9

Are you sure that this new capacitor contains a dielectric?

{snip}

 

[Zryn]

ii) C=1.2x10^6F should be C=1.2x10^-6F ... typing mistake
iii) Q = C/V should be Q = CV ... typing mistake

iv)

E=v/d ... correct
E=110/2x10^-6 ... correct
E=5.5x10^-5 J ... wrong! Be careful how you enter the numbers in your calculator. 110/2*(1x10^-6) is not the same as 110/(2x10^-6).

vi) "I transpose C=Q/V to get V=C/Q"

If you were solving for V you would get V = Q/C. And yes you are correct, when using the wrong formula (to get V= 1.51 x 10^-11), it surely cannot be right.

Clean it up and press on, you're doing rather well, just made a couple of silly mistakes.

 

[MattNotrick]

Ok guys thanks for the quick and helpfull response (is there a rep+ button?) you pointed out some mistakes for me, I've gone on and finished the question, I am unsure of a few units though. Here it is from iv)

iv)

E=v/d
E=110/2x10^-6
E=5.5x10^7

v)

E=1/2CV^2
E=1/2(1.2x10^-6)(110)^2
E=7.26x10^-3 J

vi) V=Q/C
C=(8.854x10^-12)(0.04522)/(2x10^-6)
C=2x10^-7 so
V=0.01452/(2x10^-7)
V= 72600

vii)
New capacitance = 4x10^-10 (no dielectric, thanks)
E=1/2CV^2
E=(2.4x10^-9)(72600)^2
E= 12.649824

12.649824-(7.26x10^-3)= 12.642564
Seems like quite a small change, thanks a lot for the help so far, been up late past few nights trying to do it alone

hows that so far??

 

[gneill]

Ok guys thanks for the quick and helpfull response (is there a rep+ button?) you pointed out some mistakes for me, I've gone on and finished the question, I am unsure of a few units though. Here it is from iv)

iv)

E=v/d
E=110/2x10^-6
E=5.5x10^7

Units are V/m

v)

E=1/2CV^2
E=1/2(1.2x10^-6)(110)^2
E=7.26x10^-3 J

vi) V=Q/C
C=(8.854x10^-12)(0.04522)/(2x10^-6)
C=2x10^-7 so
V=0.01452/(2x10^-7)
V= 72600

Your value for the charge looks suspicious. I recall that you calculated it correctly earlier. The units of V should be, well, V (Volts). Always put units on your results.

vii)
New capacitance = 4x10^-10 (no dielectric, thanks)
E=1/2CV^2
E=(2.4x10^-9)(72600)^2
E= 12.649824

Check your values used in the energy calculation. That doesn't look like the capacitance value you just calculated (or even half of it). The unit of Energy is J.