Question

1. Each plate of a parallel plate capacitor measures 12.0 cm by 18.0 cm. The plates are separated by a distance d = 1.80 mm of air. A 9.00 V battery is connected to the plates. The positive (+) battery terminal is connected to the positive (+) plate and the negative (-) battery terminal is connected to the negative (-) plate.
(a) What is the capacitance of the capacitor?
(b) What is the charge on the capacitor?
(c) How much energy is stored in the capacitor?
(d) What is the electric field between the capacitor's plates?

Answer 1

I referred to this link:
http://www.electronics-tutorials.ws/capacitor/cap_1.html

I found that the formula is\[c=k\left(\frac{A}{d}\right)\]

Answer 2

\[C=k\left(\frac{A}{d}\right)\]

Answer 3

what will be the answers with working and formulas

Answer 4

I think all the more I can tell you is that I think \[ k=\frac{1}{(4\pi)(\epsilon_{relative, air}\times \epsilon_0)}=\frac{1}{(4\pi)(1.00059\times 8.85418782 \times 10^{-12})}\]

Answer 5

That will let you solve for capacitance.

Answer 6

Then the charge, \(Q\), you can get from \(Q=C\ V\), which is the rearranged definition of capacitance, where \(C=\Large\frac{Q}{V}\).

Answer 7

i will try, it would be great if i get answers so i can compare to my working

Answer 8

@Festinger

Answer 9

@theEric

Answer 10

I don't know what to do about parts C and D, sorry!

All I'm certain of is that that link is probably truthful!

Good luck!

Answer 11

ima try my best lol, thanks for the help :)

Answer 12

i think capacitance of a capacitor is
c=k(A/d)
and k=aphcilon not

Answer 13

for ur question c
energy stored in capacitor is
u=(1/2)cv^2

Answer 14

now for ur question d
electric field E=v/d

Answer 15

for ur question c
energy stored in capacitor is
u=(1/2)cv^2

Answer 16

i think capacitance of a capacitor is
c=k(A/d)
and k=aphcilon not

Answer 17

Different books and people use different symbols.
Here are some holy grail of equations for Capacitors:
\[C=\frac{Q}{V}\] or
\[Q=CV\]
Which says the charge on a capacitor is the product of it's capacitance and the potential difference. The capacitance is a measure of how well the capacitor holds charge. High capacitance means for the same potential different, it can hold more charge.

Because of how Capacitance is defined, and for geometries which give nice potentials, in this case parallel plate capacitors, we can express capacitance purely by the geometry:
\[C=k\epsilon_{0}\frac{A}{d}\]
Where A is the area, d is the distance between the plates. ϵ0 is the permittivity of free space and k is something called the dielectric constant, which depends on what is between the plates. For air or vacuum it's 1.

The potential energy U stored in the capacitor is then:
\[U=\frac{Q^{2}}{2C}\]
If I substitute the equations from above Q=CV, I get 2 more forms, use whichever you prefer:
\[U=\frac{Q^{2}}{2C}=\frac{1}{2}CV^{2}=\frac{1}{2}QV\]

Since there is energy stored between the plates, I can find the energy density, which can be related by Electric field.

\[u=\frac{U}{V}=\frac{U}{Ad}=\frac{1}{2}\epsilon_{0}E^{2}\]

Now to solve. For (a):
\[C=k\epsilon_{0}\frac{A}{d}=\frac{0.0216\epsilon_{0}}{0.0018}\:F\]
F is farads, the unit for capacitance.

(b) Charge. Q. Q=CV!
\[Q=\frac{0.0216\epsilon_{0}}{0.0018}*9=9.56*10^{-10}Coulombs\]

(c) Energy! U!
\[U=\frac{1}{2}QV=9.56*10^{-10}*\frac{9}{2}=4.3*10^-9J\]

(d) E. Solve for E!
\[\frac{U}{Ad}=\frac{1}{2}\epsilon_{0}E^{2}\]

My calculations might be wrong, but the equations should be correct.

Answer 18

E should be 4999 V/m

Answer 19

Thankyou every one, i have few more questions which i will post

Answer 20

I want anyone to confirm the answers plz

Answer 21

@theEric

Answer 22

I'll look for them when you post them! And I'll help if I can.

Answer 23

2. Two horizontal conducting rails are connected by a resistor R = 0.750 Ω, as illustrated below. There is a uniform magnetic field B = 0.120 T pointing vertically downward. A conducting rod of length l = 2.20 m moves to the left along the two rails at a constant speed v = 3.90 m/s. Assume that resistance between the rod and rails is negligible.

(a) What is the induced emf in the circuit?
(b) What is the magnitude and direction of the current flow?
(c) What is the rate of heat dissipation in the resistor?
(d) What force is required to keep the rod moving at its constant speed?

Answer 24

3. An object with height ho = 4.8 mm is placed upright a distance s0 = 16 cm from the center of a concave mirror. The image is located a distance si = 7 cm from the center of the mirror.
(a) What is the focal length of the mirror?
(b) What is the size and orientation of the image?
(c) Where should the object be placed so that the image size is the same as the object size?
(d) What are the image distance and the magnification when the object is placed at a distance s0 = 4.1 cm from the center of the mirror?

Answer 25

4. A photosensitive metal is illuminated by monochromatic light with a wavelength λ = 550 nm. Electrons are emitted due to the photoelectric effect. Here are some constants that might be helpful:
1 eV = 1.60 x 10-19 J
Speed of light = 3.00 x 108 m/s
Plank’s constant = 6.63 x 10-34 J s
Mass of an electron = 9.11 x 10-31 kg
(a) If the work function of the metal is  = 1.8 eV, what is the maximum kinetic energyof an emitted electron?
(b) If the work function of the metal is  = 1.8 eV, what is the maximum speed of anemitted electron?
(c) What is the threshold (or cutoff) frequency of the metal?
(d) If the photosensitive metal is illuminated with light having a frequency equal to thethreshold frequency, what is the theoretical maximum kinetic energy an emitted electron can have?

Answer 26

I need it with working, plz plz, it is due before sunday midnight

Answer 27

@radar @Festinger @theEric

Answer 28

Please post these as separate questions so that the people working on it get credit and so that other people can see these and have a chance to answer them.

Also, I think OpenStudy permits just one question at a time, which is all we can work on anyway, @qsx .

Answer 29

ohh, so we do one by one, i will post 2nd on the after i finish first, Thank for letting me know, new user.

Answer 30

Haha, not a problem at all :)

Answer 31

And hopefully there are people who can help. So you're still working on this one?

Answer 32

right now i am doing work for a different class, i will get back to it soon

Answer 33

I see! Good luck! :)

Answer 34

Thankyou :)

Answer 35

As it has been pointed out there are different formulas you may use and @theEric has provided an excellent link which you can reference. my old books gave this formula (which I believe is the same as theEric provided:
Using this then you would calculate the capacitance as:\[(8.85\times10^{-12})(18\times10^{-2}\times12\times10^{-2})\over1.8\times10^{-3}\]\[C _{o}=106.2\times10^{-12} \]Or 106.2 pF
Did you get something like that?