Finding voltage due to spherical charge

In summary: You are given the charge density there as 1/r^2 μC/vol. So what is the total charge on the shell?the charge on the shell would be dQ=ρv*dr=1/r^2 *4πr^2*dr=4π [C] and the integral would be k*Q*∫1/r^3 dr r between 1 and 1.5 ?So, in summary, the potential difference between point A(3.0,4.0,12.0) and point B(2.0,2.0,2.0) can be found by considering the given
  • #1
eminem14
19
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Homework Statement


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)

Homework 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)

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 don't know why). Also given the subtractions shown in point 3, i couldn't integrate because the terms under the root for the magnitude of R can't be simplified.

Please help and thank u in advance
 
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  • #2
What do you know about the potential due to a uniform spherical shell?
 
  • #3
it's equal to [1/(4*pi*epsilon)] *Q / r
 
  • #4
eminem14 said:
it's equal to [1/(4*pi*epsilon)] *Q / r

Where r is what?
Then think of the given charge density as a set of such shells.
 
  • #5
r is the distance from the centre of the shell to the point charge you're trying to calculate the potential at. But i don't get how thinking of the charge density as a set of uniformly charged shells simplifies things?
 
  • #6
eminem14 said:
r is the distance from the centre of the shell to the point charge you're trying to calculate the potential at. But i don't get how thinking of the charge density as a set of uniformly charged shells simplifies things?

As long as the point for which you want the potential is outside all of the shells, the radius of each shell becomes irrelevant.
 
  • #7
so that would mean that the r - terms in the integral cancel ? and if radii are irrelevant then what is relevant besides θ & ∅?
 
  • #8
eminem14 said:
so that would mean that the r - terms in the integral cancel ? and if radii are irrelevant then what is relevant besides θ & ∅?
You've been using r to mean two different things. One is relevant, the other isn't. I'll switch one of them to s.
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 set-up, the charge density on each such shell is a function of r. Can you write down the integral?
 
  • #9
([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
haruspex said:
You've been using r to mean two different things. One is relevant, the other isn't. I'll switch one of them to s.
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 set-up, the charge density on each such shell is a function of r. Can you write down the integral?

([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<r<1.5 0 <= θ <= ∏ 0 <= ∅ <= 2∏

*the exponent of the denominator is 3/2 not 3
 
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  • #11
eminem14 said:
([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<r<1.5 0 <= θ <= ∏ 0 <= ∅ <= 2∏

*the exponent of the denominator is 3/2 not 3

You don't need any phis or thetas. You have a set of shells radius r, thickness dr. You know the volume of each shell and its charge density, so you know its total charge. You have a reference point distance s > max r from the centre of all these shells, so you know the potential at that point due to each shell. The potential due to the whole set is just the integral wrt r.
 
  • #12
haruspex said:
You don't need any phis or thetas. You have a set of shells radius r, thickness dr. You know the volume of each shell and its charge density, so you know its total charge. You have a reference point distance s > max r from the centre of all these shells, so you know the potential at that point due to each shell. The potential due to the whole set is just the integral wrt r.
so s>1.5
what would the integral be? isn't 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 ?
 
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  • #13
eminem14 said:
so s>1.5
s is whatever it is for each of these two points in turn, but it looks clear that it will be > 1.5 for each of A(3.0,4.0,12.0), B(2.0,2.0,2.0)
what would the integral be? isn't the volume of each shell would 4/3 * pi * (rout^3 - rin ^3)?
For a thin shell, you can take the volume as area * dr = 4πr2dr. You are given the charge density there as 1/r2 μC/vol. So what is the total charge on the shell?
 
  • #14
haruspex said:
s is whatever it is for each of these two points in turn, but it looks clear that it will be > 1.5 for each of A(3.0,4.0,12.0), B(2.0,2.0,2.0)

For a thin shell, you can take the volume as area * dr = 4πr2dr. You are given the charge density there as 1/r2 μC/vol. So what is the total charge on the shell?

Q would be (1/r2 * 4πr2dr)= 4∏ dr

∴ ∫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
eminem14 said:
Q would be (1/r2 * 4πr2dr)= 4∏ dr
Yes.
∴ ∫kQ/r= 1/ε∫1/r dr
You're confusing the two distances again. Try to stick to the notation I introduced: r is for the radius of a shell; s is for the distance from the centre of the shell to the point of interest.
for point A limits of the integral: 1.5 < r < s
Wrong limits. What are the min and max r for the shells?
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
Yes, it's that. Why does it seem too big?
 
  • #16
haruspex said:
Yes.

You're confusing the two distances again. Try to stick to the notation I introduced: r is for the radius of a shell; s is for the distance from the centre of the shell to the point of interest.

Wrong limits. What are the min and max r for the shells?

Yes, it's that. Why does it seem too big?

The min and max for the shells are 1.0 and 1.5. So those r the limits of the integral but then where does s fall into the integral. Is it a constant multiplied by the integral?
 
  • #17
eminem14 said:
The min and max for the shells are 1.0 and 1.5. So those r the limits of the integral
Yes.
but then where does s fall into the integral. Is it a constant multiplied by the integral?
It's the Q/r that's wrong. It should be Q/s.
 
  • #18
S is a constant though right?
 
  • #19
eminem14 said:
S is a constant though right?
For a given reference point, yes.
 
  • #20
Thanks for the help got the correct result. But I just want to understand something, why is it that the radii are irrelevant?
 
  • #21
eminem14 said:
Thanks for the help got the correct result. But I just want to understand something, why is it that the radii are irrelevant?
For a uniformly charged spherical shell, the field outside the shell does not depend on the radius of the shell. It only depends on the total charge and how far you are from the centre of the shell. There's no simple reason for this - it just turns out that way for inverse square laws operating in three dimensions. Similarly, there is no field inside the shell. So, in general, if you have a solid sphere in which the charge only depends on the distance from the centre, and you want the potential at some distance s from that centre, you can throw away all parts of the sphere further than s from the centre and consider all the charge that remains as being concentrated at the centre.
The same applies for gravitational fields. If you pretend that the Earth is uniformly dense, this gives the result that the field inside the Earth is proportional to distance from the centre.
 
  • #22
haruspex said:
For a uniformly charged spherical shell, the field outside the shell does not depend on the radius of the shell. It only depends on the total charge and how far you are from the centre of the shell. There's no simple reason for this - it just turns out that way for inverse square laws operating in three dimensions. Similarly, there is no field inside the shell. So, in general, if you have a solid sphere in which the charge only depends on the distance from the centre, and you want the potential at some distance s from that centre, you can throw away all parts of the sphere further than s from the centre and consider all the charge that remains as being concentrated at the centre.
The same applies for gravitational fields. If you pretend that the Earth is uniformly dense, this gives the result that the field inside the Earth is proportional to distance from the centre.

Thanks again and one final dumb question are the units for the voltage volts or mili volts?
 
  • #23
eminem14 said:
Thanks again and one final dumb question are the units for the voltage volts or mili volts?
You were given distances in m and charge in μC. Had the charge been in Coulombs then the answer would have been in Volts, right?
 
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  • #24
haruspex said:
You were given distances in m and charge in μC. Had the charge been in Coulombs then the answer would have been in Volts, right?

Thats what i figured. In my calculations i changed the μC to C and assumed the voltage would be in volts. Hopefully that should be correct.
 

1. What is the formula for finding the voltage due to a spherical charge?

The formula for finding the voltage (V) due to a spherical charge is V = k * Q / r, where k is the Coulomb constant (9 x 10^9 N m^2/C^2), Q is the magnitude of the charge, and r is the distance from the center of the sphere to the point where the voltage is being measured.

2. How do you determine the direction of the voltage due to a spherical charge?

The direction of the voltage due to a spherical charge is always radial, meaning it points directly away from or towards the center of the sphere. The direction is determined by the sign of the charge - a positive charge will have a voltage that points away from the center, while a negative charge will have a voltage that points towards the center.

3. Can the voltage due to a spherical charge be negative?

Yes, the voltage due to a spherical charge can be negative. This occurs when the charge is negative and the distance from the center of the sphere is greater than the radius of the sphere. In this case, the voltage will be negative and point towards the center of the sphere.

4. How is the voltage due to a spherical charge affected by distance?

The voltage due to a spherical charge is inversely proportional to the distance from the center of the sphere. This means that as the distance increases, the voltage decreases. This relationship is described by the inverse-square law, which states that the voltage is proportional to 1/r^2, where r is the distance from the center of the sphere.

5. Is the voltage due to a spherical charge affected by the presence of other charges?

Yes, the voltage due to a spherical charge can be affected by the presence of other charges. If there are other charges present, the voltage at a certain point will be the sum of the voltages due to each individual charge. This is known as the principle of superposition, which states that the total voltage at a point is equal to the sum of the individual voltages at that point.

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