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Final_HB
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Homework Statement
the separation of a parallel plate capacitor is 2mm with a potential difference between them of 45V.
(i) Find the electric field between the plates.
(ii)Find the charge per unit area of the plates if the permittivity of the dielectric medium is 20 [itex]\epsilon_0 [/itex]
(iii) Find the capacitance of each plate if the area is .13m2
(iv) If the dielectric medium between the plates is removed, so air occupies the region between the plates, find the resulting change in energy stored on the capacitor if (a) the plates are kept electrically isolated, and (b) If the plates are connected to the battery, such that the potential difference between them remains at 45V as the dielectric is removed.
(v) If the medium is only partially removed, such that half the area has the original dielectric and the other half has only air separating the plates, find the capacitance of this capacitor.
Homework Equations
Capacitance of a parallel capacitor: C= [itex]\frac{\epsilon_0 A}{d}[/itex]
Charge per unit area of capacitor: E= [itex]\frac{σ}{\epsilon_0}[/itex]
[itex]\rightarrow[/itex] σ=E ε 0
Electric field between plates: E=[itex]\frac{V}{d}[/itex]
The question says we may assume the permitivity of air is the same as a vaccum, ε
0 = 8.85x10-12
The Attempt at a Solution
The first three parts are fine. I think
(i) E= [itex]\frac{V}{d}[/itex]
E=[itex]\frac{45}{.002}[/itex]
E=22500 (units I am not sure of for an electric field
(ii) E=[itex]\frac{σ}{(20)(\epsilon_0)}[/itex] We re arrange to get σ (charge/area) on its own.
so: σ= E ε0
σ= (22500) (1.17x10^-10)
σ= 2.63x10-6 C/m2
(iii)
C=[itex]\frac{\epsilon_0 A}{d}[/itex]
C=[itex]\frac{1.17x10-10 x .13}{.002}[/itex]
C= 7.605-9 F
And here is where it gets crazy
I know that the energy stored for a capacitor is [itex]\frac{1}{2}[/itex] CV2 so since its an energy change, we find what it is originally first.
The energy stored at first then is :
E=[itex]\frac{1}{2}[/itex] (C)(V2 )
E=(.5)(7.605-9) (45)2
E=7.7x10-6 joules.
Since the dielectric has changed, C changes.
New C is just= 5.7525x10-10
Since no current is flowing in (a) The change would be -7.710-6 J (no current, no charge... right?)
And for b) Charge is going to be
(.5)(5.7525x10-10) (45)2
E=5.8244x10-7
So change in charge is:
7.11756-6
That makes sense to me anyways
And I've spend an hour trying to work out a way to start the last part, with no luck at all.
Thanks in advance
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