Recent content by Destrio

anyone?

wait, why would i get -1/2 from integrating 1/r^3 shouldnt it be -2?

ah yes, i need to brush up on those integrals, thanks! i wouldn't have been able to solve this on my own

this leaves me with the correct answer, except with 2pie on the bottom instead of 8pie (I had a factor of 2 in the numerator instead of denominator)

i can take out the 4pie as a constant so I am left with integrating dQ/r which I am not sure how to integrate.. do I want to change dQ to odA then everything is in terms of r dV = odA/r4pie = k2pirdr/(r^4)(4pie) V = int dV = (Q/4pie)(ab/b-a) * int dr/r^3 = (Q/4pie)(ab/b-a)(-2)(1/b^2 -1/a^2)...

ah yes, terrible mistake... brought over a -2 instead of a -1 so when i intergrate the final part dV = kdQ/r do i want to have dQ = odA integrating that will give me what i got for Q again and what do i do with the r?

that was what i did back when i got: k = Qab/4pi(b-a)

or will it be Q = 2pik * int dr/r^2

i figure i must be integrating dQ = odA from a to b dq = int k/r^3 dA but i know dA = 2pirdr is that correct? i don't know how to integrate that

thats not right, without the dA i end up with Q = -2k(1/b^2 - 1/a^2)

i used the - to make it 1/a - 1/b then changed it to b-a/ab so i was back at b-a the 4 i got from dA = 2pirdr then *-2 from the integral i used o=k/r^3 and odA = Q should i just use o=k/r^3 to get Q = -2k(1/b - 1/a) k = (Q/2)(ab/b-a) once i get k, what do i do with it?

or do i do: odA = dQ Q = -4pik((1/b)-(1/a)) Q = 4pik((b-a)/ab) k = Qab/4pi(b-a)

a) The current density across a cylindrical conductor of radius R varies according tot he equation j = jo(1-r/R) where r is the distance from the axis. Thus the current density is a maximum jo at the axis r=0 and decreases linearly to zero at the surface r=R. Calculate the current in terms...

so if i integrate the charge density from a to b i get -2k/r^2 Q = -2k/r^2 k = -qr^2/2 is this correct? how will i get rid of this negative?

A total amount of positive charge Q is spread onto a nonconducting, flat, circular annulus of inner radius a and outer radius b. The charge is distributed so that the charge density (charge per unit area) is giver by o = k/r^3, where r is the distance from the centre of the annulus to any point...