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The quantity of chargeq (Coulomb) pas sing through a surface of area 3 cm , changing with time as q = 4t3 + St + 6 (t is in units of seconds) . Find the current den...

Question

The quantity of chargeq (Coulomb) pas sing through a surface of area 3 cm , changing with time as q = 4t3 + St + 6 (t is in units of seconds) . Find the current density at t = 2 s a) ] = 8,50x 10+ A/m? b) ]J = 1,77X 105 A/m? c) ] = 2,83X 105 A/m? d) J = 3,94X 105 A/mz e) ] = 5,08X 105 A/m?

The quantity of chargeq (Coulomb) pas sing through a surface of area 3 cm , changing with time as q = 4t3 + St + 6 (t is in units of seconds) . Find the current density at t = 2 s a) ] = 8,50x 10+ A/m? b) ]J = 1,77X 105 A/m? c) ] = 2,83X 105 A/m? d) J = 3,94X 105 A/mz e) ] = 5,08X 105 A/m?



Answers

The quantity of charge $q$ (in coulombs) that has passed through a surface of area 2.00 $\mathrm{cm}^{2}$ varies with time according to the equation $q=4 t^{3}+5 t+6,$ where $t$ is in seconds. (a) What is the instantaneous current through the surface at $t=1.00 \mathrm{s}$ ? (b) What is the value of the current density?

In this question, we are given ah the charge that passes through the surface. Uh You know, as a function of time. Take two is 4.00. teach you less. I. P 5.00. T. Plus 6.00. You want to find a current uh to the surface. That he goes to one second and the value of the current density at the same time. Okay, so to do this question in a he'll be using I go through the TTP. So uh current as a function of time is the beauty of 40- 96. Okay. Did make sure you get trapped t square last five. So T. E. goes to 1/2. Okay. The current is 12 past five because to 17 mps. So this is the answer for a then happy we want to calculate their current density. So jay goes too high over a. Okay. Well, given that the current is 17, the area is 2cm square. Um so we convert to meters where two times 10 to the negative Bob and the answer is 8.5 times 10 to the power for I am pure per square. Okay, so this is the answer poppy and I saw for this question.

So for this problem were given the area and they were given the amount of charge that passes through at each time. T So for part A, we want to find the instantaneous current and has to be equal to de que over d t. So we want to take the derivative of this equation with respect to time. And we find that that is equal to 12 t squared plus five just taking the power rule. And then if we plug in t equals one, we get that the current is equal to 12 times one squared plus five. So 17 amps, then for part B, we're looking for the current density, which is just the current over the area. We have the current we saw from part A and in the area that were given in the problem. So Jay z me equal to 8.5 times 10 to the four amps per meter squared

Okay, so in this problem, we have a box. So I sports here. Arbaugh are books. This is a really bad books. Let's do it again. Okay. Ah, what else? And we want to calculate in the first part What is the volume? The density volume charge at the position that's put here in green of the position. Uh, 0.3. That is correct. Little 0.3 meters. Okay, so So we know that the electric field is describe here. This is the electric field line and here is entering in one side of the box, Um, leaving on the other side of the box. The IV area vector off the surface off the letter of the boxes is going to be one pointing here and the order pointing in here. Okay. Therefore, to calculate the charge density charge that we want, we must look, we must remember about the electric flux, and we know that the electric flux is described us the electric shoot multiply by the area motor vibrate because sign off teeter, which is also equal the charge inside divided by absolute zero. OK, but the charge inside in this problem is going to be the density that we want to calculate. That multiplies the volume off the box divided by absolute zero. No, we can calculate what we want. So let's look to desecration. Here we have that the electric feud there multiplies the area multiplies because sign off, Teeter is going to be equal hope feet divided by absolute zero. But there is only one problem here. We want the net electric flux. And this year is not in that electric flux. This is the flux from one of the surface. So the natural electric flux is going to be something like this. Ah, is going to be the electric flux through the first face Miners, the electric flux through the second face. Okay, not minus plus. Sorry. So the net flux is the flux. True, the face one. We lost the flux through the face to, and then we can use desecration. So the equation that we must so is going to be simply fight to just the electric future. Um, electric fields there multiplies the area. One the multiplies because sign off 1 80 plus the electric field, the multiplies the area to they're multiplies the core sign off zero degrees and that is equal. Ho v divided by absolute zero. Okay, so now miscalculate what we actually want. So we know that the electric field is constant. Electric feud is constant. And let me see what is the area off. Okay, so we know that the electric field is express us six times 10 to the tree or to simplify is just 600 multiplies the area, the multiplies, the ex the first X. Let's call the X one is square. Go sign off 1 80 which is negative. So that's what Here. Uh, negative. X one square, los X two square, Go side of zero. Richest one. This is equal to Hope V absolute zero. Okay, so who is the position? X one is 0.3, as we can see in the initiate of the problem. So this is just 600. A de multiplies negative. Um, zero point three square. Tell us 0.3. Pull us the X square. This is a product that we must. So So this is going to be zero point three square loss to actually, uh, plus zero point six g x. Tell us DX is square. Okay, which is equal to hold V absolute zero. So if we neglect the D X Square because it's too small is a differential so small that we can neglect We only have 0.3 bless 0.6 d x. So as we can see, we have negative 0.3 plus 0.3 can cross this one and we can simply five or equation to just 600. Hey, they're multiplies 0.6 dx, which is equal to hope. Who is the volume? The volume is just a d x, the area times the differential element divided by absolute zero. Then we can see that we can cross the area we can cross the D X and our density charge that we want to calculate is going to be just ho equals, uh, 600. That multiplies 0.6 that multiplies absolute zero. So this is just actually now that I see, I forgot the zero here. This is not 600. We have six to the power, but six times tend to the tree. This is 6000. Sorry, guys. I forgot to zero here a zero here and a zero here. Because of that, we also forgot a zero in here. 6000 and Knox is a 600. Okay, now that is correct. If remote supply all this, we know that absolute zero is going to be equal eight point 85 for two times tented a minus 12. Therefore, we have density. Charge off three points. 18. 19. Sorry. Group or 19 times tend to the miners. Eight cool apps per meters. Q Bix, That's the answer to the first part and the second bart way. We must answer if this region ah, could be inside a conductor and well, the answer is no, because the field will have to be zero. Because it's so small, it have to be zero. Therefore, bestie, final answer. Thanks for watching, Okay?

In this question, we have a charge which has a uniform volume density equal to 1.2 nana coolant per meter cubed as indicated here by row. We know that it feels an infinite slab between the X coordinates minus five and plus five. What we want to find the magnitude of the electric field at a point with the cord and the X equals four and at any point with the cord and the X equals six centimeters. So what we know is that X is not point not four m. The net field acts right would from the charge lying between minus not 40.5 and not point not four and left word on the other side. So in the region from not point not four to not point not five m, we know that sigma. So the total charge is equal to the total enclosed charge over the area. As stated by Gaza's Law. Where this is the charge density, we also know that this is equal two row times the volume over a which is also equal to road delta X in the case of this question. So what we can do is we can say that the net electric field mm indicated by the module is to say that it's the magnitude is equal to the contribution on the left side, subtracting the contribution on the right side, so right, no point not nine m over to epsilon not minus because this part of the forces acting in the opposite direction row not point not one over to absolutely not. Where this is the distance between minus not point not five and the position of where we're trying to find out the field. So this is a difference of point not nine m and meters. And here is the distance between X is not point not four so four centimeters and X is not point not five so five centimeters. So plugging in the values we've got we can say that this is equal to 1.2 times 10 to the minus nine which is row multiplied by not point not nine minus not point not one. Yeah. Yeah. All over to epsilon. Not where episode or not, has a value of 8.85 times 10 to the minus 12. Yeah. Okay. Yeah. Yeah. This gives us a final answer for where X equals four centimeters of the force of magnitude 5.4. Newton speculum. The second part of this question is when X is equal to six centimetres in this case the field contribution from all layers of the charge point. Right words. So in this case we would add these distances. So we can say that the magnitude of the electric field is equal two row times not 20.1 as this is not point not nine plus, not point not 1/2 X. L. Or not, which is equal to 1.2 times 10 to the minus nine. Mhm times not 0.1 over two times. Absolutely not again, which has the value again of a constant of 8.85 times 10 to the minus 12, which gives us a final answer in this case of 6.8 newton speculum.


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