Two 2.00 cm×2.00 cm plates that form a parallel-plate capacito...

Question

Two 2.00 cm×2.00 cm plates that form a parallel-plate capacitor are charged to ±0.708 nC. What are the electric field strength inside and the potential difference across the capacitor if the spacing between the plates is 1.00 mm?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem. Two parallel plates form a capacitor with a plate area of 12.0 centimeters squared and a separation distance of 1.20 millimeters. The plates are charged to a net charge of plus or minus 0.708 micro columns. I calculate the electric field strength between the plates. Note assume even electric field distribution between the plates. I I find the electric potential difference between the plates. OK. So we have two different answers that we're trying to find. So our end goal is to find an two different answers we need to find for I, we need to find the electric field strength between the plates. And then our second answer we need for part I I, we need to find the electric potential difference between the plates. Awesome. We're also given some multiple choice answers for part I and for part I I, so let's read them off to see what our final answer pair might be A for I is 66.7 mega volts per meter and I I is 80 kilovolts B for I is 96.0 mega volts per meter. And I I is 53.3 kilovolts C for I is 96.0 mega volts per meter and I I is 80 kilovolts. And finally, for D for part I is 66.7 mega volts per meter and I I is 0.12 mega volts. Awesome. So first off, let us assume that the electric field between the plates is uniform. Now we can start to solve. For part I let us recall and use the fact that a charge on plate, Q is related to the electric field. E so E represents the electric field and that is equal to a divided by epsilon subscript zero, which is also equal to Q divided by capital A all divided by epsilon subscript zero. And let's quickly define our variable. So ada represents the area charge density epsilon subscript zero represents the permittivity of free space constant and capital A represents the surface area of the plate. And don't forget that capital Q is the charge on the plate. OK. So we're gonna use that the electric field is equal to Q divided by A, all divided by epsilon subscript zero. So we'll take this equation and we need to simplify it and solve for E. So when we do that. So when we simplify, we get that the electric field is equal to Q divided by capital a multiplied by epsilon subscript zero. But before I start f like solving for this, let's quickly off to the side here, make a little side note what's note that one Newton per Coolum is equal to one bolt per meter. And this is going to be very important when we are canceling out our units and simplifying units. This will make more sense in a second. OK. So now we can plug in all of our known variables and sulfur E to get our answer for part I. So let's do that. Let's write it down below. So we don't run out of room. So E is equal to. So our value for Q is given to us in the prom itself as 0.708 micro Coolum. But we need to convert micro Coolum to Coolum. So all we need to do is multiply this by 10 to the power of negative six. Now we need to divide that by our area multiplied by the permittivity of free space constant. So the area, so the surface area of the plate is given to us as 12 centimeters squared, but we need to convert centimeters squared to meters squared. So let's do that really quick. So 12 0.0 centimeters squared multiplied by, let's recall that in 1 m there is 100 centimeters. So 1 m divided by 100 centimeters squared. So when we multiply that out, the centimeters squared will cancel out leaving us with meters squared, which is exactly what we need. And that's our area. And now we need to multiply it by the permittivity of free space constant which the numerical value for that is eight point 85, multiplied by 10 to the power of negative 12 cool LS squared divided by newtons multiplied by meters squared. Awesome. OK. So when we take this big old long equation and we plug it into a calculator, we will find that it will equal six point 67, multiplied by 10 to the power of seven newtons per Coolum. But let's remember that one newton per Coolum is equal to one volt per meter. So that's the same as saying 6.67 multiplied by 10 to the power of seven volts per meter. Awesome. But as we can tell all of our multiple choice answers are in mega volts per meter. So we need to take 6.67 multiplied by 10 to the power of seven. And we need to divide it by 10 to the power of six. So we'll get 66.7 mega volts per meter. And that is our final answer. For part I, we did it awesome. So now we could start solving for part I I, so let's do that. So to solve for part I I, we must recall and use the equation for the relationship between the electric potential difference, which is represented by delta v and the electric field represented by E and the plate separation represented by D for the capacitor. So we can write that equation as the electric field cap represented by capital E is equal to delta v, which is the electric potential difference divided by the separation between the plates which is represented by lower case D. So now we need to rearrange this equation to solve for delta V. So we need to isolate and get delta V by itself. So when we do that, we'll get that delta V is equal to capital E multiplied by D. So the electric potential difference is equal to the electric field multiplied by the plate separation. So now we can plug in all of our known variables to solve for delta V. So let's note that are electric field value in this case will be the value we found in part I but we don't need, we don't want to use the mega volts per meter value. We need to use the volts per meter value. So we need to put write down 6.67 multiplied by 10 to the power of seven volts per meter. And we need to multiply it by the separation or plate separation. I should say which the plate separation was 1.20 millimeters, but we need to convert millimeters to meters. So let's do a quick conversion. Let's recall that in 1 m there is 1000 millimeters. Awesome. So the millimeters cancel out leaving us with meters, which is exactly what we want. So when we plug, listen to a calculator, we will find that delta V is equal to 80040. So 80,040 which is the same as in scientific notation as 8.0 multiplied by 10 to the power of four volts. But as you can see, none of our multiple choice answers are in volts. So let's convert our answer to Kila volts. So all we have to do to convert it to Kila volts is we take 8.0 multiplied by 10 to the power of four and we divide it by 1000. And when we do that, we will get 80 kg volts, which is our final answer. For part I I hooray, we did it. Awesome. So for part I, we found that the electric field is equal to 66.7 mega volts per meter. And then for part I, I, we found that delta V, the potential difference was 80 kilovolts. Fantastic. So let's look at our multiple choice answers to see which answers are final answer and the correct answer. And it turns out that her answer has to be the letter A. So I is 66.7 mega volts per meter and I I is 80 kilovolts. Thank you so much for watching. Hopefully that helps and I can't wait to see you in the next video. Bye.

Two 2.00 cm×2.00 cm plates that form a parallel-plate capacitor are charged to ±0.708 nC. What are the electric field strength inside and the potential difference across the capacitor if the spacing between the plates is 1.00 mm?

Key Concepts

Electric Field in a Capacitor

Capacitance

Potential Difference

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