Electrical Physics | Solving Problems with Flux and Nonconducting Shells

[rjnara]

Hey Everyone,
I had the great idea to enroll in this physics as a distance ed. course. I quickly have discovered that I am supposed to understand everything just by reading the text. Anyhow I have come across some problems that whatever I do I can't figure out so here it goes:
1. Flux and nonconducting shells.
A charged particle is suspended at the center of two concentric spherical shells that are very thin and made of nonconducting material. Figure 23-31a shows a cross section. Figure -1-31b gives the net flux through a Gaussian sphere centered on the particle, as a function of the radius r of the sphere.
Now, the diagram illustrates that at the point of interest, the net flux is -4*10^5 and asks what the charge of ring "A" is. To solve for this I tried many things: 1. using the equation qenc=Flux*epsilon*naut (didn't work), Next I tried subtracting the central charge from the net Flux*epsilon*naut at the point of interest. Neither of these worked.

2.Two charged concentric spherical shells have radii of 10.0 cm and 15.0 cm. The charge on the inner shell is 4.10*10-8 C and that on the outer shell is 2.50*10-8 C. Find the magnitude of the electric field at r=11.0 cm and r=20.5 cm.
Honestly I didn't know where to start, so I just looked at the formula in the book E=8.99*10^9*(q/R^3)*r. Where q=4*10^-8, R=10cm and r=11cm.
I really appreciate any help with this.

 

[Andrew Mason]

Hey Everyone,
I had the great idea to enroll in this physics as a distance ed. course. I quickly have discovered that I am supposed to understand everything just by reading the text. Anyhow I have come across some problems that whatever I do I can't figure out so here it goes:
1. Flux and nonconducting shells.
A charged particle is suspended at the center of two concentric spherical shells that are very thin and made of nonconducting material. Figure 23-31a shows a cross section. Figure -1-31b gives the net flux through a Gaussian sphere centered on the particle, as a function of the radius r of the sphere.

Now, the diagram illustrates that at the point of interest, the net flux is -4*10^5 and asks what the charge of ring "A" is. To solve for this I tried many things: 1. using the equation qenc=Flux*epsilon*naut (didn't work), Next I tried subtracting the central charge from the net Flux*epsilon*naut at the point of interest. Neither of these worked.

We will need either Figure 1-31b or its contents in order to help you answer this question.

2.Two charged concentric spherical shells have radii of 10.0 cm and 15.0 cm. The charge on the inner shell is 4.10*10-8 C and that on the outer shell is 2.50*10-8 C. Find the magnitude of the electric field at r=11.0 cm and r=20.5 cm. Honestly I didn't know where to start, so I just looked at the formula in the book E=8.99*10^9*(q/R^3)*r. Where q=4*10^-8, R=10cm and r=11cm.
I really appreciate any help with this.

I can see why you are confused. The answer given is not right.

This appears to be just an application of Gauss' law or Coulomb's law, if we assume that the charge is evenly distributed over the area of the two shells.

$$\int E\cdot dA = E*4\pi r^2 = q_{encl}/\epsilon_0$$

Or, by application of Coulomb's law:

$$E = kq/r^2$$

For the first part,

$$E = q_{1}/4\pi\epsilon_0r^2 = kq_1/r^2$$

For the second,

$$E = (q_1 + q_2)/4\pi\epsilon_0r^2 = k(q_1+q_2)/r^2$$

AM

 

[kyle8921]

Hello there,
It's great that you are taking the initiative to learn about electrical physics through a distance education course. It can be challenging to understand complex concepts just by reading the text, so don't get discouraged! Let's address the two problems you mentioned:

1. Flux and nonconducting shells:
From the given information, we can see that the net flux through the Gaussian sphere is -4*10^5. This means that there is a negative charge enclosed within the sphere. To find the charge of ring "A", we can use the formula qenc = Flux * epsilon_0, where qenc is the enclosed charge, Flux is the net flux, and epsilon_0 is the permittivity of free space. However, since the Gaussian sphere is centered on the particle, the charge of the particle will also contribute to the enclosed charge. Therefore, we need to subtract the charge of the particle from the result we get from the formula. This should give us the charge of ring "A".

2. Two charged concentric spherical shells:
To find the electric field at a given point, we can use the formula E = k * q / r^2, where k is the Coulomb's constant, q is the charge, and r is the distance from the charge. For the first point (r=11.0 cm), we can use the formula with the charge on the inner shell (4.10*10^-8 C) and the distance of 10cm. For the second point (r=20.5 cm), we can use the formula with the charge on the outer shell (2.50*10^-8 C) and the distance of 15cm. Remember to convert the distances to meters before plugging them into the formula.

I hope this helps you with your problems. If you are still having trouble, don't hesitate to reach out to your instructor or fellow classmates for assistance. Keep up the good work!