a. $103 \mathrm{A} / \mathrm{m}^{2}$b. $1.16 \times 10^{13} \mathrm{V} / \mathrm{m} \cdot \mathrm{s}$c. $1.30 \times 10^{-6} \mathrm{T}$d. $0.650 \times 10^{-7} \mathrm{T}$




You are watching: A parallel-plate air-filled capacitor is being charged as in the figure

*

05:38


*

Salamat A.



*



See more: (G)I-Dle I Am

*

Video Transcript

mm in this problem. First step is to start up for what she know. So the given information we know the radius of the parallel plate capacitor Ambassador is four centimeters converting that into meters. Yeah 0.04. We also know that the conduction current. I see conduction current Is equal to 0.5- zero. And here for part they were told to find the displacement current density. So for part A we have part A ask yourself what is the displacement current density with JB. In the air space between those between the plates. So displacement current density is given that J. D. Is equal to I. D. Over eggs. Where I. D. IBM the conduction current and A. Is the area of the capacitor. So A. The area which is high art square. We could figure out what the radius is Given in the problem which is 0.04. Where put that into your calculator. You have 0.0050. There are points zero zero. I have to say me this script. Now we can find our JV. So J. B. Your I. D. Zero point 0.5-0 divided by your area. Zero point 00 5 to 6. Your J. B. becomes 98 points 85 for part the barbie is what is the rate at which the electric field between the plates between the place? It's changing. So the rate of change of an electric field is given by the E over G. T. So the rate of change of their electric of the electric field of respect to time. So the E over D. C. Is equal to J. B. Over absalon zero. You know JD displacement current density & F. Salon zero is with the glen in the primitive itty of free space. So absolutely zero. Its value is Given us eight 854 tom's tent in a good call. We know that J. D. S. So we could find a G E D. T. It's 98 0.85 but Divided by 8.85. 4 Times 10 to the negative. 12 input ellis your calculator. This becomes one point Home playing Mom 1 6 Times 10 to the -13 times 10 to the 13 part C. We're told to figure out the magnetic field between the plates At a distance of 22.00 cm on the axis. So they induced magnetic field between the place. Can be found using Ampere Law. And we know ampere law is magnetic field B is equal to New zero over to sons. JB times are NEW. Sub zero is the approach is the U. Sub zero's Um use um zero is the probability a free space and its value is equal to four pi times tense in the making of seven. And we know what JD is JD is displacement current density. And our is the distance from the act. So B is equal to four pi you know what baby is 98.85 times are. And the R. Is two centimeters So time 0.02. So and put them into your calculator, input into your calculator. You get one point 242 Times 10 to the -6. And magnetic field is measuring Tesla. So T magnetic field density is measuring tougher Bacardi. What told what is induced magnetic field between the plates at one Month cm from the Axis. So magnetic field density be equal to he was zero over to time. Baby times are. Yeah four pi Tom's tense and a negative seven over to times 98 0.85 Time 0.01. And put them into your calculator. You should get six point 210 Times 10 to the -7 test. And there you have