Calculating Capacitor Potential Difference and Resistance for 12V Battery

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The discussion revolves around calculating the potential difference and resistance in capacitor and resistor scenarios involving a 12V battery and a 9V battery. For the capacitor filled with a dielectric, participants clarify that the energy stored can be equated using the formula U = 1/2 CV^2, leading to a derived potential difference of approximately 5.66V for the dielectric-filled capacitor. The resistance of a resistor connected to a 9V battery is calculated using the energy delivered over time, with participants confirming the relationship between power, energy, and resistance. There is a consensus that capacitance cancels out in the equations provided, simplifying the calculations. Overall, the thread emphasizes the importance of understanding energy storage in capacitors and the relationship between voltage, resistance, and energy in resistive circuits.
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Two equipotential surfaces that surround a +3.0 x 10^-7 C point charge has a radius of 0.15m, what is the potential surface

Two capacitors are identical, except that one is empty and the other is filled with a dielectric (k= 4.50). The empty capacitor is connected to a 12.0 V battery. What must be the potential difference across the plates of the capacitor filled with a dielectric such that it stores the same amount of electrical energy as the empty capacitor?

A resistor is connected across the terminals of a 9.0-V battery which delivers 1.1x10^5 J of energy to the resistor in six hours what is the resistance to the resister? (i solved this one but got a ridiculously small resistance of like 3.4 to the negative 8)
 
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2nd one:

E = 1/2CV^2

E1 = 1/2C x 144

E2 = 4.5 x 1/2 x C x V = 1/2C x 144

Ummm... don't know if that's right, but it looks like it. I mean, kappa basically mutliples the capacitance. C's cancel.

3rd one: My way of thinking is: P = V^2/R

J = V^2/R x T (In your case 3600) You know everything but R, so plug and chug.

Sorry, forgot how to do the first one. We went over this for like a day, then went on to circuits =S

PL
 
For the third one, http://www.sengpielaudio.com/calculator-ohm.htm.
Your first question is unclear.
For the second one, can you tell me the energy in the capacitor without the dielectric? Qualitatively, do you expect the required voltage to be more or less than the case with k=1?
Do you have a textbook?
 
Would it matter what the capacitance is? Doens't it just cancel out anyway?

And don't give him (her?) those links. :p He'll (she?) never learn that way. :)

PL
 
the second question is exactly as it is out of the book, and i don't have the book, i left it at school. thnx for helping guys, i appreciate it... =P. i have a chem 102 exam tmmrw morning at 8 am too... i am screwed.
 
crap is Watts=J/s?
 
Completed.
A resistor is connected across the terminals of a 9.0-V battery which delivers 1.1x10^5 J of energy to the resistor in six hours what is the resistance to the resister? (and i figured it out without the link =) i found out that P is J/s hahaha rather than just joules. haha stupid watts. thnx for helping out.)

Thnx a lot guys only 2 to go.

Hrmm... i don't actually know if capacity matters but i think it probably does...
 
Poop-Loops said:
Would it matter what the capacitance is? Doens't it just cancel out anyway?

And don't give him (her?) those links. :p He'll (she?) never learn that way. :)

PL
The capacitance will of course cancel since we are given no information about the capacitor itself.
The energy store in a capacitor is U = \frac{1}{2} CV^2.
The problem at hand states \frac{1}{2} C_1V_1^2= \frac{1}{2} C_2V_2^2. You are looking for V_2.
 
i believe V works out to be 5.66 =) thnx guys.
 
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