## anonymous 3 years ago 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?

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1. theEric

I referred to this link: http://www.electronics-tutorials.ws/capacitor/cap_1.html I found that the formula is$c=k\left(\frac{A}{d}\right)$

2. theEric

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

3. anonymous

what will be the answers with working and formulas

4. theEric

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

5. theEric

That will let you solve for capacitance.

6. theEric

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

7. anonymous

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

8. anonymous

@Festinger

9. anonymous

@theEric

10. theEric

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!

11. anonymous

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

12. shamim

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

13. shamim

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

14. shamim

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

15. shamim

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

16. shamim

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

17. Festinger

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.

18. Festinger

E should be 4999 V/m

19. anonymous

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

20. anonymous

I want anyone to confirm the answers plz

21. anonymous

@theEric

22. theEric

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

23. anonymous

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?

24. anonymous

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?

25. anonymous

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?

26. anonymous

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

27. anonymous

28. theEric

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 .

29. anonymous

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

30. theEric

Haha, not a problem at all :)

31. theEric

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

32. anonymous

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

33. theEric

I see! Good luck! :)

34. anonymous

Thankyou :)

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:|dw:1374787084437:dw| 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?

If you are pursuing an Electrical Engineering career you have a lot of work to look forward to. It will be worth it. Good luck with it.

37. theEric

@radar I read your "About Me" thing, and it's all cool and admirable! Is that an area of electrical engineering? Also, is $$\epsilon_0$$ used only for free space between the plates?

38. theEric

yes air, if someother dielectric is used like Mica, then you adjust by multiplying by 5 and use $\epsilon _{r}$$\epsilon _{r}=5\epsilon _{o}$

@theEric I never obtained a college EE degree, but I converted to an Electronics Engineer in my government career, becoming classified as GS 855-**. Years prior I was an electronic tech GS-856 Those numbers are U.S. government classifications of job titles. Most of the time I worked with radar. But have retired and don't do much anymore.

I am going to check out those links you posted

I noticed that the latter link gave Mica a 3-6, my old text gave it a 5, (which is within the range.

@theEric I just got through reading your profile, I would of guessed you were going for an EE, but computer science does include the hardware (or at least I think it does) and I believe you will become involved in the electronics area.

44. anonymous

Thank yuo :)

45. anonymous

after all working i ended up with this (a) The capacitance of the capacitor C = OA/d Where, O=8.85*10-12 , A=Area of plate , d=distance between plates C = 8.85*10-12*(18*12)*10-4/(1.8*10-3) = 1.062*10-10 F (b) Charge on the capacitor Q = C*V = 1.062*10-10 *9 = 9.558 *10-10 C = 9.558*10-4 C (c) Energy is stored in the capacitor = *C*V2 = * 1.062*10-10 *92 = 4.301*10-9 =4.301*10-3 J (d)The electric field between the capacitor plates = V/d = 9/(1.8*10-3 ) = 5*105 V/m

46. anonymous

@Festinger

47. Festinger

E=5000 V/m

48. anonymous

(a) C=ε 0 A /d =$8.85x10^{-12}$x0.0216/.0018=106.2 pF, (b) Q=9X106.2 pC=955.8pC (c) U = $1/2 C V ^{2}$ = 1/2 X 106.2* 81 pJ=4301.1 pico Joule (d) E = V/ d = 9 /.0018 = 5000 N/C

49. anonymous

@Fifciol

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