A parallel-plate capacitor has capacitance C0 = 8.00 pF when there is air between the plates. The separation between the plates is 1.50 mm. (a) What is the maximum magnitude of charge Q that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00x10^4 V/m? (b) A dielectric with K = 2.70 is inserted between the plates of the capacitor, completely filling the volume between the plates. Now what is the maximum magnitude of charge on each plate if the electric field between the plates is not to exceed 3.00x10^4 V/m?

Video transcript

Welcome back everybody. We are taking a look at a camera flash unit that uses a parallel plate capacitor inside. Now we are told that there are no di electrics between the capacitors. Right. Um, so with that being said you uses a parallel plate capacitor that is 150 micro fare adds or 100 and 50 times 10 to the negative six Fair adds the distance between the two metal plates of the capacitor is 20. m three millimeters and we are told that the electric field between them is six times 10 to the fourth per meter. Now we are tasked with finding two different things. We are tasked with finding first what is the maximum charge allowed on the capacitor And then to say we do introduce a teflon dielectric to the system With a constant of 2.1. What will then be the new maximum charge for the capacitor? Let's go ahead and start with part one here. What is our maximum charge of the capacitor as it stands? Well, this is just going to be the initial capacitance, times the voltage. But what is the voltage shared here or voltage is just going to be the electric field times the distance between them. So we have six times 10 to the fourth times 100.0 three giving us a voltage of 180 volts. Now we can find our maximum charge which will be equal to 50 times 10 to the negative six Fair adds times 180 volts giving us a maximum charge of 2. times 10 to the negative second. Cool ums great. So now moving on to part two here we have that a teflon. Dielectric is introduced meaning that we are going to be able to calculate a new maximum charge. Now we're going to use the same formula except now with the dielectric there's going to be a new capacitance. So we have to calculate that first. Our new capacity is just going to be equal to our constant that we are given times our old capacitance. So this will be 2.1 times 1 50 times 10 to the negative six. Giving us a new capacitance of 3.15 times 10 to the negative fourth fair ads with that. Let's go ahead and find our new maximum charge. This is going to be equal to 3.15 times 10 to the negative fourth times 1 80 volts giving us 5.67 times 10 to the negative second columns. So we found the maximum charge without a dielectric. We found the maximum charge with a dielectric and this corresponds to our final answer choice of C. Thank you all so much for watching. Hope this video helped. We will see you all in the next one