A charge of -6.50 nC is spread uniformly over the surface of one face of a nonconducting disk of radius 1.25 cm. (d) Why is the field in part (a) stronger than the field in part (b)? Why is the field in part (c) the strongest of the three fields?

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Hi everyone in this particular practice problem, we are asked to actually compare the electric field on a disc ring and a sphere where we'll have a thin nonconducting disk, a nonconducting ring and a tiny sphere. Each of them carrying a charge of 130 Nano coom uniformly distributed over their services. The charge on the disc is located on one face, the disk and the ring have the same radius of 10 centimeters and each electric field, the E disc or the electric field on the disc is the field produced at point X one located on the axis of the disc. And at a distance of 30 centimeters from its center. And the electric field on the ring or E ring is the field produced at point X two located on the axis of the ring and at a distance 30 centimeters from its center and E E sphere or the electric field of the sphere is the field produced at a point B 3, 30 centimeters away from the sphere. We were asked to compare the electric field in the disc during and the sphere. Which one is the biggest? Which one is the second biggest and which one is the smallest. So first, what we wanna do is to probably tackle the uniform disk. So for the disc at a point of X equals 30 centimeters or X one, um I'm just gonna say X one equals 30 centimeters like so, and the disc itself has the radius of 10 centimeters, which is um 0.1 m, but we're just gonna leave it this uh in centimeters from now. So at point uh X one located on the axis of the disc and at a distance of 30 centimeters from its center of the disc itself, the electric field can be calculated following this formula right here. We wanna recall that the electric field of a disc can be calculated by sigma over to epsilon knot. Uh parenthesis, open parenthesis, 1 -1 over the square root of one plus parenthesis R over X squared, close bracket or close parenthesis. And this right here can actually be um further derived into Q over two pi R squared epsilon knot. So the sigma here can be described into Q over A or Q over pi R squared. And whatever's on the parenthesis will remain the same, which is 101 minus one over the square root of one plus R over X squared just like. So OK, so we actually have all the necessary information to calculate this. And we can actually just plug in all the values right now. The Q is known to be 130 nano Cullom uniformly distributed over the surface of the ring. So that will be 1 30 times 10 to the power of minus nine Coulon just um converting from nano Coolum to Coon over two pi is still gonna be two pi and then the R squared is remember that we want to convert our R here two m. So that's this is going to be 0.1 m squared. And then the uh epsilon knot is a constant which is gonna be 8.85 times 10 to the power of minus 12 Coolum per oops Coolum per Newton meter squared, just like. So and whatever is inside of the parenthesis is still going to be the same. I am going to remove or I'm going to uh Kind of switched us to the left side a little bit to give us more space and still -9. And then whatever's on inside of the parenthesis will remains the same. And this is going to be the square root of one plus R. The R is 10 centimeters. But in this case, we can just uh let it be in centimeters because the X which is X one is actually known in centimeters as well. So the two units in centimeters will cancel out. So that is going to be OK to use in our calculation. So I'm just gonna write down 10 centimeters over 30 centimeters squared and all of this after plugging it into our calculator will come out to be a value of electric field disc to be 1.20 times 10 to the power of four newton per. So that will be the answer for the electric filk of the disc. OK. Now, we want to move on to calculating the electric field for the ring. So for the ring, we will have the same um radius which is um R of 10 cm. I'm gonna indicate this one with a big r now to make it a little bit different, But we still have the same point which is X or the same distance of the point of measurement which is X2, which is going to be 30 cm from the uh axis of the ring or from the center of the ring. So for a ring, we can actually recall the formula to calculate the electric field, which is going to be 1/4 pi epsilon knot multiplied by Q X over X squared plus the radius squared to the power of 3/2. And this X right here is going to be X two. OK. So now we can actually um um input the formula that we have. So the 1/4 pe knot is actually a known constant, we can input all of our known values. Now. So the known constant of 1/4 pi epsilon knot will be 9.0 times 10 to the power of nine Newton multiplied by meter squared over Coolum square. And this will be multiplied by the Q or the charge which is 1 nano coom which N Coolum is 1 30 times 10 to the power of minus nine Coolum. And then the X, we want it to be in si, we have 30 centimeters converting that to SI is 0.3 m. And then the denominator on the bottom is X two squared or this is also X two sorry X two squared, which is 0.3 m squared plus R squared, which is 10 centimeter which is 0.1 m squared Just like so to the power of 3/2. So after plugging all of this into our calculator, we will have E ring two B 1.1 10 oh 1.10. Then then to the power of four Newton per K. OK. So now that we have the E ring, the last uh value that we need is the uh electric field for the sphere. So the field produced at a point P 3, 30 centimeters away from the sphere will be. So for the sphere um And I'm gonna say P three equals 30 cm. So E sphere will equals to one over. We want to recall the formula for calculating the electric field of a sphere. Uh This will be 1/4 pe knot multiplied by Q over R squared. And this formula of uh this constant of 1/4 pe knot will be the same, which is nine times 10 to the power of nine Newton multiplied by meter squared over kum squared. And then the Q will be the same which is 130 nano column which is 1 30 times 10 to the power of minus nine colum N si and the R square is 0.3. Or in this case, the R here is going to be our P T value which is the distance where the electric field is measured. So I'm gonna probably just change it to P three to make it uh easier for us to keep track. So this is going to be BT squared and this is going to be 0.3 m squared which is in si and this will give us a value of the E sphere of 1.3. I stand to the power of four Newton per colon. So 1.3 times 10 to the power of four newtons per colon just like. So OK, so now that we have concluded all the E or all the values for the electric field on the disc during and the sphere, we can compare them on to one another. So as we can see, the E disc is 1.2 times 10 to the power of four net newton per colum. The E ring is 1.1 times 10 to the power of four newton per column. And the E sphere is 1.3 times 10 to the power of four newtons per column. So, oops, I'm gonna write down our conclusion here. So is fear is going to be larger than E disc and going to also be larger than E E ring where E E ring is going to be the smallest. And that will be the answer to our problem which will actually correspond to option A. So in this case, option A is going to be the answer to our problem and that will be it for this particular practice problem. Um If you guys have any sort of confusion on this, please make sure to check out our other lesson videos on similar topics and that will be all for this video. Thank you.

A charge of -6.50 nC is spread uniformly over the surface of one face of a nonconducting disk of radius 1.25 cm. (d) Why is the field in part (a) stronger than the field in part (b)? Why is the field in part (c) the strongest of the three fields?

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Key Concepts

Electric Field

Surface Charge Density

Superposition Principle