Finding voltage due to spherical chargeby eminem14 Tags: electromagnetics, potential, potential difference, spherical charge, vector calculus 

#1
Jul2613, 09:06 PM

P: 19

1. The problem statement, all variables and given/known data
A volume charge density of ρv=1/r^2 μC/m^3 exists in the region bounded by 1.0m<r<1.5m. Find the potential difference between point A(3.0,4.0,12.0) and point B(2.0,2.0,2.0) 2. Relevant equations dQ=(ρv)dv dV=[dQ/(4∏ε0]*norm(R) where R is position vector of a point charge relative to observation point A or B. dv=r^2 *sinθ*dr*dθ*d∅ x=rsinθ*cos∅ ; y=r*sinθ*sin∅ ; z=r*cosθ (transformations of coordiantes from spherical to cartesian Ar= Ax*sinθ*cos∅ + Ay*sinθ*sin∅ + Az*cosθ (obtaining radial component(Ar) from cartesian components Ax,Ay and Az) 3. The attempt at a solution I attempted to first find the voltage at A and B and subtract them from each other by doing the following: 1. Finding dQ=[1/(r^2)]*r^2 *sinθ*dr*dθ*d∅=sinθ*dr*dθ*d∅ 2. Therefore, dV=sinθ*dr*dθ*d∅*norm(R) 3. Rx= r*sinθ*cos∅ 3 Ry=r*sinθ*sin∅ 4 Rz=r*cosθ 12 (trasformation of spherical components to cartesian coordinates minus the coordinates of A. Problem: I am finding difficulty in getting the norm of R because I can't obtain the unit vector of R which means I can't evaluate the integral to find V. I assumed that the voltage would only have a radial component given the method used in the matlab solution I have(not sure though and dont know why). Also given the subtractions shown in point 3, i couldnt integrate because the terms under the root for the magnitude of R can't be simplified. Please help and thank u in advance 



#2
Jul2613, 11:09 PM

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What do you know about the potential due to a uniform spherical shell?




#3
Jul2613, 11:31 PM

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it's equal to [1/(4*pi*epsilon)] *Q / r




#4
Jul2713, 01:20 AM

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Finding voltage due to spherical chargeThen think of the given charge density as a set of such shells. 



#5
Jul2713, 02:22 AM

P: 19

r is the distance from the centre of the shell to the point charge you're trying to calculate the potential at. But i dont get how thinking of the charge density as a set of uniformly charged shells simplifies things?




#6
Jul2713, 02:31 AM

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#7
Jul2713, 11:53 AM

P: 19

so that would mean that the r  terms in the integral cancel ? and if radii are irrelevant then what is relevant besides θ & ∅?




#8
Jul2713, 04:47 PM

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Consider a shell radius r, thickness dr, and a point distance s > r from its centre. You correctly gave the formula for the potential at that point as [1/(4*pi*epsilon)] *Q / s. In your setup, the charge density on each such shell is a function of r. Can you write down the integral? 



#9
Jul2713, 06:42 PM

P: 19

([10^6]/(4*∏*ε)*∫∫∫[itex]\frac{sinθ*dr*dθ*d∅}{(r*sinθcos∅  3)^2 + (r*sinθ*sin∅ 4)^2 + (r*cosθ  12)^2)^3/2}[/itex]
where the limits of the integral: 1. 1<r<1.5 2. 0 <= θ <= ∏ 3. 0 <= ∅ <= 2∏ 



#10
Jul2713, 06:47 PM

P: 19

where the limits of the integral: 1<r<1.5 0 <= θ <= ∏ 0 <= ∅ <= 2∏ *the exponent of the denominator is 3/2 not 3 



#11
Jul2713, 08:37 PM

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#12
Jul2713, 09:22 PM

P: 19

what would the integral be? isnt the volume of each shell would 4/3 * pi * (rout^3  rin ^3)? and what about the coordinates of A & B? where do they fall into this if Q can be calculated would the integral be: k*Q*∫1/r^3 dr r between 1 and 1.5 ? 



#13
Jul2813, 01:37 AM

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#14
Jul2813, 10:00 AM

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∴ ∫kQ/r= 1/ε∫1/r dr for point A limits of the integral: 1.5 < r < s and ∫kQ/r= 10^6/ε∫1/r dr for point B limits of the integral: 1.5 < r < s but how is s calculated for each point? is it the magnitude of position vector pointing to the point → mag(3i + 4j + 12k ) = root(169)= 13 seems too big 



#15
Jul2813, 04:40 PM

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#16
Jul2813, 04:45 PM

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#17
Jul2813, 06:45 PM

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#18
Jul2813, 06:54 PM

P: 19

S is a constant though right?



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