Question

An air-filled parallel-plate capacitor has an area of 1.5 m^2 and a capacitance of 1 nF.

(Questions below.)

Answer 1

a) What is the distance between the plates? Express your answer using three significant figures. Answer in mm.

b) If the capacitor is connected to a 50 V battery, what is the electric field inside the plates? Answer in N/C.

c) If the capacitor is connected to a 50 V battery, what is the magnitude of the charge on each plate? Answer in C.

d) When the capacitor is connected to a 50 V battery, what is the energy stored in the capacitor? Answer in J.

e) The battery is disconnected from the capacitor so that the charge stays on each plate. You push the plates inwards so that the distance between the plates is half of what it was. What is the energy stored on the plates now? Answer in J.

f) How much work (positive or negative) are you doing on the plates moving them inwards? Why does this work make sense in terms of the force and displacement you apply? Answer in J.

Answer 2

a)
$$
C=\varepsilon_r\varepsilon_0\frac{A}{d}\\
\varepsilon_0=8.854×10^{−12} F/ m\\
\varepsilon_r=1\\
A=1.5~m^2\\
C=1.5\times10^{-9}~Farads\\
d=\cfrac{\varepsilon_r\varepsilon_0A}{C}~m
$$
http://en.wikipedia.org/wiki/Capacitance
b)
$$
E=\cfrac{V}{d}~Volts/m
$$
Where d is given in part d.
c)
$$
C=\cfrac{Q}{V}\\
Q=CV~Coulombs
$$
d)
$$
U=\cfrac{1}{2}CV^2~Joules
$$
http://en.wikipedia.org/wiki/Capacitor#Energy_of_electric_field
e)
$$
\large{
E_{new}=\cfrac{V}{d/2}\\
V_{new}=\cfrac{dE_{new}}{2}=\cfrac{1}{2}E_{old}\\
U_{new}=\cfrac{1}{2}C\left (\cfrac{1}{2}E_{old}\right )^2~Joules\\
=\cfrac{1}{8}CE_{old}^2=\cfrac{1}{2}U_{old}
}
$$
So, the energy (U) between the places has decreased.
f)
As part d showed, if you move the plates closer, there is less energy between the plates. Therefore, energy was lost through the work to move them inward. Therefore, you did negative work, because the plates did positive work, i.e it lost energy.

To quantify, take the difference between part e and part d above.

That the plates did positive work makes sense because the plates are attracted to each other and therefore the force and the direction of displacement are in the same direction.

Answer 3

a) d=((Er)(Eo)(A))/C
=((1)(9*10^-12)(1.5))/(1.5*10^-9)
=0.009m > 9 mm

Hm... I got this one wrong.

Answer 4

d=eoA/C
= ((8.89*10^-12)(1.5)) / (1*10^-9)
= 13.3 m

^ that gave me the right answer

Answer 5

c) Q=CV
=(1*10^-9)(50)
= 5 * 10^-8

Got that right

Answer 6

b) E=v/d
= 50/0.0133
= 3760

Got that

Answer 7

d) U=(1/2)CV^2
= (0.5)(1*10^-9)(50^2)
= 1.25*10^-6

Got that one too

Answer 8

e) Just half of d)?
2.5*10^-6 is wrong, so I guess I'm misunderstanding it.

Answer 9

0.0133/2=0.00665
((8.85*10^-12)(1.5))/(0.00665*10^-3) = 2.0*10^-6

(0.5)(2.0*10^-6)(50^2) = 0.0025

Ugh, I don't get this one x.x

Answer 10

0.0133/2=0.00665
((8.85*10^-12)(1.5))/(0.00665*10) = 2.0*10^-9

(0.5)(2.0*10^-9)(50^2) = 2.5*10^-6

Oops, I already changed it to m.
I'm just getting the same thing as before?

Answer 11

Oh, I got it, never mind.
I wasn't halfing it right.

0.625*10^-6 and then it's just negative for f)

Ahah, thanks!