What is the radial electric field at the point r = 2R2? Give your answer in units of kN/C.
I think there needs to be more information here. Don't you?
yes i did not know to how insert it after i copied the question. but here is the other part.
A hollow non-conducting spherical shell has inner radius R1 = 9 cm and outer radius R2 = 19 cm. A charge Q = -35 nC lies at the center of the shell. The shell carries a spherically symmetric charge density ρ = Ar for R1 < r < R2 that increases linearly with radius, where A = 17 μC/m4.
So it looks like it gives you a charge and a spherical shell of charge, and then asks for the electric field strength a certain length away from the center charge. Does that sound right to you? I'm tired, so I'll have to think about this one. The electric field at that point will be the sum of the electric fields of all the charges. So the center charge is simple. The shell of charges will have to be treated as it suggests, where there is a density of charge throughout dependent on \(r\). And we'll probably integrate to account for the electric field of the overall shell.
So, the center charge is easy, right? Do you know the formula for the electric field at a point due to one charge?
i found out a final equation for the first part but when i go and plug in my given info its wrong, my equation was this: E=4K/(R1+R2)^2)(Q+A*pi((R1+R2)/2)^4-R1^4)
Did you derive your equation, or find it somewhere? I guess we should make sure it's correct. Does it account for the center charge of -35nC and separation of \(2R_2\)?
Did you use a double integral?
a classmate and i derive it, but no it was a single integral, we started it like this...E*A=4*pi*K*Qenc where Qenc=(Q+integra of dq)
that was for the first part but for the second part i started the same but changed the integral dq to pdv
I have to get going in 10 minutes, just so you know. I'm wondering how to use the polar coordinate system to calculate the separation between any point in the shell and some point \(2R_2\) away from the center.|dw:1396820558603:dw| I think it would be a double integral, since the separation depends on the distance from the center and the angle. In my picture, I labeled them \(r\) and \(\theta\) respectively.
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