Two 2.0 cm×2.0 cm metal electrodes are spaced 1.0 mm apart and...

Question

Two 2.0 cm×2.0 cm metal electrodes are spaced 1.0 mm apart and connected by wires to the terminals of a 9.0 V battery. The wires are disconnected, and insulated handles are used to pull the plates apart to a new spacing of 2.0 mm. What are the charge on each electrode and the potential difference between them?

Accepted Answer

Welcome back. Everyone in this problem. Two conducting carbon rods of diameter, 1.5 centimeters are connected to a 15 volt battery. The circular phases of the rods are parallel and separated by three millimeters forming a capacitor. The rods are disconnected from the battery when the current stops flowing after the disconnection insulators are used to alter the rod separation to four millimeters. And we want to calculate the charge and the potential difference between electrodes after the separation. For our answer choices. A says that our charge is 7.8 times 10 to the negative 12 coulombs and our potential difference is 11 volts B says it's 7.8 times 10 to the negative 12 Coulombs and 20 volts C says it's 5.9 times 10 to the negative 12 Coulombs and 11 volts. And D says it's 5.9 times 10 to the negative 12 coulombs and 20 volts. Now, here we're talking about two possible scenarios. Let's try to represent them on a diagram. Now, in the first scenario, we're talking about a carbon rod with a diameter of 1.5 centimeters where the phases of both rods are parallel and separated by three millimeters to form a capacitor. So we could say that it probably looks something like this. Let me represent it here. So it's a parallel play. But you know, but let me, let me try to make sure to show in a way that it's circular. I know, let me play it safe and enjoy it like this. OK? So we know that this is before or let's say that this is initially initially, right? So initially, we have our separation of three millimeters, OK? For our conducting carbon rods with a diameter of 1.5 centimeters there, we also know that both of them are connected to a 15 volt battery. So we could say then that we have our voltage source here, that's 15 volts and the same on the other side, I I won't put it on it because the word initial is here. But I think you get the idea now in the second scenario, OK? I think I can show it in the second scenario. It's in the finance, let's call that scenario, the final scenario, we have our capacitors, but now the separated insulation has been added to make them have a separation of four millimeters and they're disconnected from the battery. And now we want to calculate the charge and the potential difference between the electrodes after the separation. So basically, that's what we're interested in. So in other words, we want to find the final charge, OK? And the final voltage call that VF now what do we know about charge? But in a parallel plate capacitor to recall that the charge is given by the capacitance multiplied by the potential difference. This implies then that the potential difference equals the charge divided by the capacitance. No, the thing is the charge between the or the charge remains the same even after the increase in separation. So if we can find the initial charge, it will be the same as the final charge. OK. So the final charge will be equal to the initial charge. So let's first see if we can figure out what our initial charge will be. Already, we know that our potential difference is 15 volts. Both of them are connected to a 15 volt battery source. OK? But we don't know what their capacitance is going to be. We don't know what their capacitance is going to be there in this case, the initial capacitance. So let's see if we can figure that out, what else do we know about capacitance? Well, before the increase in separation between the circular phases of the rods, the capacitance. So now we're solving for our initial capacitance in the initial capacitance would be equal to the permittivity of free space multiplied by the area of the plate divided by or the area of our rod rather divided by our initial distance of separation. OK. And we know that to figure out what the area is for R ra we can, the area of a circle is pi R squared. So we can rewrite that as enon multiplied by pi R squared divided by the initial separation. Now, we know that uh the permittivity of free space as a is a constant. And we also know that the initial distance is three millimeters. And we can also tell that the radius, we know that the diameter is 1.5 centimeters. So let, let me write that here. We know that our radius is going to be half of 1.5 centimeters, which is 0.5 centimeters. So what does that tell us then? Well, since we have all that information, we can solve for our initial capacitance and thus solve for our initial charge. So substitute that substituting that to solve for our charge. OK. Solving for our charge does remember the initial charge equals the final charge. Then our charge is going to be equal to initial charge is gonna be equal to our initial capacitance multiplied by the change or the potential difference. OK. Which using what we have here is going to be equal to ek not multiplied by pi R squared divided by the initial separation multiplied by the potential difference. And now substituting all our values here. OK. That means it's going to be equal to 8.85 times 10 to the negative 12 square coulombs per Newton square meters multiplied by pi multiplied by our radius squared, which is 0.75 centimeters, spin negative 2 m all squared multiplied by our potential difference which is 15 volts. And all of that, all of that is going to be divided by our initial separation, which is three millimeters. OK. So that will be our initial charge. No, what do we get when we work? That's all. Well, let me put this into our calculator. OK. Then this tells us, this tells us, this tells us that our charge, our initial charge, our final charge doesn't matter because the charge doesn't change even if the spacing between the rods is increased. So our final charge really is going to be equal to 7.8 times 10 to the negative 12 coulombs. So that's the charge between the electrodes after the separation. So now we found our final charge now that we found our final charge, we just need to find the potential difference after the increase in separation. So let's put that in blue here. So now we're solving for delta V fine. OK. So solving for that again, let's recall, what do we know about our potential difference? Recall that the potential difference then is going to be equal to the charge divided by the final capacitance. The final potential difference equals the charge divided by the final capacitance. So what do we know about the final capacitance? Well, it's the same plate. So we know it's the same area but no, the distance uh has changed the the separation distance has changed from three millimeters to four millimeters. So with that in mind, then we already have the formula for our capacitance. So that means our final capacitance is going to be equal to E knot multiplied by pi R squared divided by our final separation. So now if we substitute that into our formula, then this tells us that our final potential difference is going to be equal to a charge divided by OK. Divided by E knot pi R squared, divided by the F which let me make sure my division sign is showing properly which then is going to be equal to Q multiplied by our final separation divided by E kno pi R squared. OK. So now when we put that, let's not substitute the values because we know all of those values. So our charge is 7.8 7.8 times 10 to the negative 12 full arms. OK? And that is going to be multiplying our final distance which is four millimeters four times 10 to the negative 3 m. OK? And all of that is being divided by E knot 8.85 times 10 to the negative 12 square coulomb per Newton square meters. OK? Multiplied by pi multiplied by R squared. So again, that would be 0.75 times 10 to the negative 2 m all squared. OK? No, we have our formula. We, we, we know all the values to find the potential difference, the final potential difference. So let's put that into our calculator. Now, when we do that, we should find that our final potential difference, that is the potential difference between the electrodes after the separation is approximately equal to 20 votes. So the charge after the separation is 7.8 times 10 to the negative 12 coulombs. And the potential difference after the separation is 20 volts. When we look back on our answer choices, that would have been answer, choice, answer choice. B thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Capacitance

Electric Field

Charge Conservation

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