Two 10-cm-diameter electrodes 0.50 cm apart form a parallel-plate...

Question

Two 10-cm-diameter electrodes 0.50 cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 15 V battery. After a long time, the capacitor is disconnected from the battery but is not discharged. What are the charge on each electrode, the electric field strength inside the capacitor, and the potential difference between the electrodesc. After the original electrodes (not the modified electrodes of part b) are expanded until they are 20 cm in diameter?

Accepted Answer

Hey, everyone. So let's see what this question is asking us. We have two solid discs with a radius of our knot is equal to 12 centimeters and a separation of eight millimeters that form a parallel plate capacitor. An 18 volt power supply is connected to the disks. After an infinitely long time, the wires are disconnected while the charge on the disk is conserved. After disconnecting the power supply, the radius of the disks is changed to 1.5 or not. We're asked to calculate part one, the charge stored on the disks. Part two, the electrical field strength between the discs and three. The potential difference between the electrodes at the separation of 1.5 times or not. Our multiple choice answers are a 0.9 nano columns, one kilovolt per meter and eight volts B two nano columns, 2.3 kilovolts per meter and 18 volts C two nano columns, 2.3 kilovolts per meter and eight volts or D 0.9 nano equals one kilowatt per meter and 18 volts. So to solve part one of this problem, we can recognize that because the charge on the disks is conserved, we can calculate the charge at are not rather than at that 1.5 R knot. And so the equation for the charge on parallel plate capacitors is given by Q is equal to V are voltage multiplied by a, our area multiplied by epsilon knot, the electric constant divided by D our distance. And so plugging in the values that we know our voltage is 18 volts. Our area area of a disc or a circle is pi R squared. So pi multiplied by the radius 12 centimeters or 0.12 m squared multiplied by our electric constant, which we can recall is 8.85 times 10 to the negative 12. And that's cool MS per volt meter all divided by our distance. That's between there, that's the distance of separation. And so eight millimeters can be rewritten as eight times 10 to the negative 3 m. So we plug that in and we get nine times 10 to the negative 10 coulombs which equals 0.9 nano coons. And so looking at our multiple choice answers, we can eliminate B and C based on our answer to problem or to part one of the problem. Now, for part two, we can recall that our electric field strength or E is equal to our charge density denoted by sigma divided by epsilon knot. And in turn, that charge density is simply Q divided by area. So we have our electric field strength is equal to Q or charge divided by a or area multiplied by epsilon knot. And so we can use the charge, we just found nine times 10 to the negative 10 columns divided by the area. So we have pie and now RR is going to be 1.5 times of our initial R. So we're at that 1.5 R knot. So we have 18 or centimeters or 0.18 m squared multiplied by our electric constant, 8.85 times 10 to the negative 12 columns per volt meter plug that in and we get our electric field strength of one kilovolt per meter. And part three asks us to find the electric potential difference. And the key to that is recalling the equation V is equal to e multiplied by D. So our electric potential difference is going to be equal to one K volt per meter. So one times 10 to the three volts per meter multiplied by that distance separating the two discs, which was eight times 10 to the negative 3 m and that comes out to eight volts. And so when we look at our multiple choice answers, we can see that answer. Choice. A is the correct answer. So that's all we have for this one. We'll see you in the next video.

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