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Parallel-plate air capacitor of area A 14.7 cm2 and plate separation of d 2.00 mm is charged by battery to voltage of 52.0 V. What is the charge on the capacitor? 3...

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Parallel-plate air capacitor of area A 14.7 cm2 and plate separation of d 2.00 mm is charged by battery to voltage of 52.0 V. What is the charge on the capacitor? 3.40*10"-7 What is the capacitance of parallel plate capacitor? Submit Answer Incorrec: Tries 2/40 Previous Triesdielectric material with K 4490 is inserted so that it fills the volume between the plates (with the capacitor still connected to the battery) , nOW much additional charge will flow irom the battery onto the positive pl

parallel-plate air capacitor of area A 14.7 cm2 and plate separation of d 2.00 mm is charged by battery to voltage of 52.0 V. What is the charge on the capacitor? 3.40*10"-7 What is the capacitance of parallel plate capacitor? Submit Answer Incorrec: Tries 2/40 Previous Tries dielectric material with K 4490 is inserted so that it fills the volume between the plates (with the capacitor still connected to the battery) , nOW much additional charge will flow irom the battery onto the positive plate? Submit Answer Tries 0/40 Post Discussion Send Feedback



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A parallel-plate capacitor with plate area $A=2.0 \mathrm{m}^{2}$ and plate separation $d=3.0 \mathrm{mm}$ is connected to a $45-\mathrm{V}$ battery (Fig. 40 $\mathrm{a} )$ . (a) Determine the charge on the capacitor, the electric field, the capacitance, and the energy stored in the capacitor. (b) With the capacitor still connected to the battery, a slab of plastic with dielectric strength $K=3.2$ is placed between the plates of the capacitor, so that the gap is completely filled with the dielectric. What are the new values of charge, electric field, capacitance, and the energy $U$ stored in the capacitor?

In this question you have a parody update capacitor. Okay, So it's connected to a six volt battery. Right? So in this case what we have over here is uh huh. Initial energy start in 72 zero Nano Jews. Then after you insert the dielectric. Yeah. Oh yeah. Okay. Uh I'll tell you is equal to 317 you know jews. Okay. So therefore passing this question, we want to find a dielectric constant. The charge on the please after the dielectric is inserted and uh mental the feud between the plates before and after the electric is inserted. Okay, so to do by a find a dielectric constant. Okay so we know that. Yeah dielectric. Okay see prime is kappa psi not your prime is uh huh. Have c prime be square which is half coppa seen on you're square. And it's this papa you know, okay, time you're not are you i is equal to 72, you know Jews, are you crying is equal to 72 plus 3 +17 Which is China 89 9 of Juice. And copy it would just be your prime over. You know Which is China 89 divided by 72. And then you get 5.40. Okay, so this is a dielectric constant. Okay. And part B. We need to find out charge saw on the uh capacitor after the directory. Who has been inserted? So you're using U. Equals to have QB. Mhm. So Q. Would just be too you divide by B. Because here you, is there Your prime in primate? A. So two times 3 89. Uh You know, jews get by by six words and you get 130, you know, columns. Okay, so this is the answer for puppy. The next I see you want to find a electrocute before the dielectric has been inserted so we can find this ah Finding A Cure 1st. Before the dielectric, wow. Uh is inserted. Okay to not this to you. Not divide by B. Just two times 70-90 jewels. Given by six votes. And you get 24 and columns. Okay. So for paralympic, What parallel pain arrangement? Two yes. Equal to sigma or X. Or not? Which is Q. Over A. X. Or not? So and since we have given the area we can compete directly. So 24 times 10 to the -9. Goodbye. Barry. 50 times 10 to the -4. 8.5 times center to limit yourself. He and you probably do as you get 5.42. Thanks 10 to the five votes power meter. Yeah. Okay so this is the answer for Patsy. He for paddy. Want to calculate the E. The electric field after the dielectric is inserted. So I think so. Capacitor, he's still connected to the veggie after the battery is inserted. E after Is equal to e. four. Okay. And you get 5.52. Thanks 10- five. Sports car. Uh huh. So this is the answer for paddy. Mm. That's all for this question.

So we're given a parallel plate capacitor with an area of 2.3 centimeter squared. I want to convert that two meters. So we're going from centimeters squared, two meters squared. So we have 100 squared centimeters squared in one squared meters, guardia per square, everything. And this goes this 2.3 times 10 to the negative fourth meters me know that the plates of the capacitor are separated by 1.5 millimeters. Den conferred that two meters. We have 1000 millimeters in one meter. So that gives this one point five times 10 to the negative. Three meters of this one should be squared because this is an area that's a distance. And now we want to know the capacitance of this capacitor. And since we know it's air filled, then its capacitance has to be the area of the plates times Absalon, not the electrical primitive ity of free space divided by the distance separating the plates so well to play All these numbers area we found is 2.3 times 10 to the negative fourth meters squared. Absalon not is 8.85 times 10 to the negative 12th and the distance separating them. It's 1.5 times 10 to the negative three. Okay, this gives us a capacitance of 1.36 times 10 to the negative 12th Farage's, which can also be written as 1.36 PICO ferrets. Okay, so now the capacitor is connected to a 12 volt battery. So we have our voltages, 12 bowls and we want to know the charge. Will the charges given by CV we found see in the last part is 1.36 times 10 to the negative 12th and we know the voltages 12 volts. So this gives us a charge of 1.63 times 10 to the negative 11 cool ums, which is also 16.3 PICO cool ums. And now, for the last part, we want to know the magnitude of the electric field. So the voltage between any two plates as long as the field this uniform is just going to be the field times the distance. So the field is volts per distance. We have 12 volts and our distance is 1.5 times 10 to the negative three. So this gives us a field between the plates of 8000 volts per meter

But a of the Union problem. Sebastian's is given by K. Times 1 8 times aided by Betty Okay. is one times 8.85 times 10 to the power -12 Times the area we have is 2.30 times 10 to the power -4. Divided by a distance of 1.54 times since our -3 m. This gives us capacitance of 1.36 peak of arad. But to be the charges story will be called to capacitance times Chairman potential. The bastions is uh 1.36 peak of around that is in federal. It These 1.36 times gentle power -12 Time. The potential difference of 12. This gives us charges of 16 0.3 people. Cool number. From here we can find the electric field. I'd like to feel this game by ah the change in potential Distance. The change in potential is 12 divided by distance 1.5 times 10 to the power -3. This gives us an electric field of 8000 rules per meter

But a of the given problem. The capacitance is given by K Times Sloan Times A divided by D Is the area so welcome Ward Given Centimeter Square Idea. A two meter square in one ply the constantly by the distance. One times 8.85 times 10 to the minus 12 times. 2.30 Time Stanford Bar minus four Divided by 1.5 old time Stamp. Our minus three. This gives us capacitance of 1.36 people. Fare out. Pardon me. The Q amount of jewelry store will be called the capacitance Times. Change in potential. The best ins we have is 1.36 times 10 to the power, minus 12 times the potential difference of troll this game's discharge off 16.3 people. Fair art eso, Equal said. Aren't cool or but see the electric field. Is he cool, too? Potential difference. Do you like me? Distance Potential difference. We have 12 instances. One point fivefold downstairs. Bauer minus three. This gives us an electric field off 1000 rules for a meeting


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