How Does Gauss' Law Apply to Spherical Charge Distributions?

In summary, a charge of 2pC is uniformly distributed throughout the volume between concentric spherical surfaces having radii of 1.3 cm and 3.3 cm. The magnitude of the electric field 1.8 cm from the center of these surfaces is 9.02 N/C.
  • #1
Punchlinegirl
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A charge of 2pC is uniformly distributed throughout the volume between concentric spherical surfaces having radii of 1.3 cm and 3.3 cm. What is the magnitude of the electric field 1.8 cm from the center of these surfaces? Answer in units of N/C.

I used the equation [tex] \Phi = E*4\pi r^2 [/tex] and then the equation [tex] \Phi = q_e_n_c /E_0 [/tex] and set them equal to each other. I used the formula q_e_n_c = charge/volume * volume to get [tex] Q/(4/3)\pi R^3 * (4/3)\pi r^3. [/tex] Simplifying gave me [tex] Qr^3/E_0 R^3 [/tex]. Plugging this in for q_e_n_c and solving for the E field gave me ,
[tex] E= Qr/ E_0R^3*4\pi [/tex]
Q= 2 x 10^-12 since the charge is uniform
r =.018 m
E_0 = 8.85 x 10^-12
R= .033
So 2 x 10^-12(.018)/8.85 x 10^-12 *(.033)^3 * 4pi = 9.02 N/C... which isn't right.
Can someone please help me? I have no idea what I'm doing. :confused:
 
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  • #2
The electric field depends on the enclosed charge, and from the problem statement, it is the charge enclosed within the radius of 0.018 m, of the charge distribution between 0.013 m and 0.033 m. In other words, the charge enclosed in a shell of inner radius 0.013 m and outer radius 0.018 m.

2 x 10^-12(.018)/8.85 x 10^-12 *(.033)^3 * 4pi
implies the charge 2 pC distributed over the entire spherical volume of radius 0.033 m.

The 2 pC is distributed uniformly in a spherical shell 4/3[itex]\pi[/itex]((0.033 m)3 - (0.013 m)3).

Then determine what charge is enclosed by the smaller spherical shell of inner radius 0.013 m and outer radius 0.018 m.
 
  • #3
Ok.. I'm not sure if I am doing this right but...
[tex] q_e_n_c = 2 x 10^-12 * (.013)^3 / (.018)^3 [/tex]. Solving this would give me 7.53 x 10^-13 which is my Q. Then I plug this into my previous equation, [tex] Qr /E_0 R^3 4\pi [/tex]
Where Q= 7.53 x 10^-13
r= .013
R = .018
Is this the right way to do it?
 
  • #4
Think of how one finds the area between two circles, and by extension, the volume between two spherical surfaces.
 
  • #5
Ok so to find the area between the 2 circles, I would used [tex] 4\piR^3 - 4\pi r^2 [/tex] since they're spheres. Then I would multiply this answer by E for the equation for flux. I'm a little confused about the volume. Would I have [tex] Q/ (4/3)\pi (R^3 - r^3) * 4/3 \pi r^3 [/tex] ? I understand that I need to find the difference in volume between the 2 spheres, but I'm a little unsure of where it would go. Thanks
 
  • #6
Please be careful.

The area between two circles is simply the difference in areas,

A2 - A1 = [itex]\pi\,{r_2}^2\,-\,\pi\,{r_1}^2[/itex] = [itex]\pi\,({r_2}^2\,-\,{r_1}^2 )[/itex], with obviously r2 > r1

and then likewise, the volume between two spheres is simply the difference in volumes,

V2 - V1 = [itex]4/3\,\pi\,{r_2}^3\,-\,4/3\,\pi\,{r_1}^3[/itex] = [itex]4/3\,\pi\,({r_2}^3\,-\,{r_1}^3 )[/itex], again obviously with r2 > r1
 
  • #7
I'm sorry, I typed my previous post wrong. Anyway, I tried starting over using the help you gave.
[tex]\Phi = E*A [/tex]
[tex] \Phi= E * \pi (R^2 - r^2 [/tex]
where R= .018 and r= .013
So [tex] \Phi= 1.55 x 10^-4 \pi * E [/tex]
Then [tex] \Phi= q_e_n_c / E_0 [/tex]
[tex] Q_e_n_c = charge/volume * volume [/tex]
[tex] Q_e_n_c= 2 x 10^-12 / (4/3)\pi (.033^3 -.013^3) * (4/3)\pi (.018^3 -.013^3) [/tex]
So [tex] Q_e_n_c = 2.16 x 10^-13 [/tex]
Then [tex] 1.55 x 10^-4 \pi *E = 2.16 x 10^-13/ 8.85 x 10^-12 [/tex]
Solving for E gave me 50.1.
Am I doing this right now?
 
  • #8
[tex] Q_{enc}= 2 x 10^-12 / (4/3)\pi (.033^3 -.013^3) * (4/3)\pi (.018^3 -.013^3) [/tex] is correct,

however, I would write it

[tex]Q\,\frac{0.018^3-0.013^3}{0.033^3-0.013^3}[/tex]

in order to show clearly which is the numerator and which is the denominator. :smile:
 
  • #9
Is the first part of my flux wrong? It doesn't matter how I write the charge, I'm still getting 50.1 as my answer and it's wrong. I don't know how it can we wrong now!
 
  • #10
Can someone please tell me what I'm doing wrong?
 

FAQ: How Does Gauss' Law Apply to Spherical Charge Distributions?

What is Gauss' Law and how does it relate to charges?

Gauss' Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the total charge enclosed within that surface. In other words, it describes the relationship between electric charges and the electric field they produce.

What is the mathematical expression for Gauss' Law?

The mathematical expression for Gauss' Law is ∮SE⃗ · dA⃗ = Qenc0, where ∮SE⃗ · dA⃗ represents the electric flux through a closed surface S, Qenc is the total charge enclosed within that surface, and ε0 is the permittivity of free space.

How can Gauss' Law be applied to calculate the electric field of a charge distribution?

Gauss' Law can be used to calculate the electric field of a charge distribution by choosing an appropriate Gaussian surface that encloses the charge distribution. The electric flux through this surface can then be calculated using the mathematical expression for Gauss' Law, and the electric field can be determined by solving for E⃗.

What is the difference between a point charge and a continuous charge distribution in Gauss' Law?

A point charge is a single, isolated charge, while a continuous charge distribution refers to a collection of charges that are distributed over a non-zero volume. In Gauss' Law, the electric field due to a point charge can be calculated by considering the charge as a point source, while the electric field due to a continuous charge distribution requires the use of integration to consider all the individual charges that make up the distribution.

Can Gauss' Law be used to calculate the electric field of a non-uniform charge distribution?

Yes, Gauss' Law can be applied to calculate the electric field of a non-uniform charge distribution. However, the calculation may be more complicated and require the use of calculus to integrate over the non-uniform charge distribution. In some cases, it may be more efficient to use other methods, such as Coulomb's Law, to calculate the electric field of a non-uniform charge distribution.

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