Finding the potential between two charged conducting shells?

In summary: Your Name]In summary, the proper method for calculating the potential at a point inside two charged shells of different radii involves using integrals. By integrating the formula for electric potential over the entire region between the two shells, we can find the total potential at that point. This method is more accurate than simply summing the potentials of the individual shells and is the preferred method by most scientists.
  • #1
JustinLiang
81
0

Homework Statement


Suppose we have an outer shell with charge -2Q and radius b. We also have an inner shell with charge Q and radius a. what is the potential when a<r<b.


Homework Equations


E=kQ/r^2
V=kQ/r
E=-dV/dx

The Attempt at a Solution


The method I have been using according to my professor is a mathematical coincidence and the proper method to solving this is to use integrals.

How I solve it is that I sum the potential of the outer shell with the inner shell. So since we are inside the -2Q shell we have a constant V=-2Q/b. Since we are outside the Q shell we have V=Q/r.

Thus the resulting potential is V=Q/r-2Q/b.


I would like to learn the proper method where you have to take infinity as a reference and integrate. And then you get +C somewhere... lol Could someone please explain this? Thanks!
 
Physics news on Phys.org
  • #2


Thank you for your question. The method you have been using is indeed a mathematical coincidence and not the proper way to solve this problem. As you mentioned, the correct method involves using integrals.

To start, we can use the formula for electric potential, V=kQ/r, to calculate the potential at a point inside the inner shell (a<r<b). We can then integrate this over the entire region to find the total potential at that point.

The integral for electric potential is given by V=-∫E•dr, where E is the electric field and r is the position vector. In this case, the electric field is given by E=kQ/r^2, so we can rewrite the integral as V=-∫kQ/r^2•dr.

Since we are solving for the potential at a point between the two shells, we can set up our integral as follows: V=-∫a^b kQ/r^2•dr. Note that we are taking the limits of the integral as the inner and outer shell radii, a and b, respectively. This is because we are integrating over the entire region between the two shells.

Integrating this expression gives us V=-kQ∫a^b 1/r^2•dr. The integral of 1/r^2 is -1/r, so we can rewrite the expression as V=kQ(1/a-1/b). This is the proper way to solve for the potential at a point between two charged shells.

I hope this explanation helps you understand the proper method for solving this problem. If you have any further questions, please do not hesitate to ask. Thank you for your interest in science and for striving to learn the correct methods for solving scientific problems. Keep up the good work!
 

1. What is the formula for finding the potential between two charged conducting shells?

The formula for finding the potential between two charged conducting shells is: V = kQ/r, where V is the potential, k is the Coulomb's constant, Q is the charge on one of the shells, and r is the distance between the two shells.

2. How do I determine the charges on the two conducting shells?

The charges on the two conducting shells can be determined by using the principle of charge conservation. This means that the total charge on one shell must be equal and opposite to the total charge on the other shell for the overall system to be electrically neutral.

3. Can the potential between two charged conducting shells be negative?

Yes, the potential between two charged conducting shells can be negative. This occurs when the charges on the two shells are of opposite signs and the distance between them is large enough. In this case, the potential will be negative, meaning that the electric field points from the shell with a positive charge to the shell with a negative charge.

4. How does the distance between the two shells affect the potential?

The distance between the two shells has an inverse relationship with the potential. As the distance increases, the potential decreases. This is because the electric field weakens with distance, resulting in a lower potential between the two shells.

5. Can the potential between two charged conducting shells be affected by other external charges?

Yes, the potential between two charged conducting shells can be affected by other external charges. This is because external charges can create an electric field that interacts with the electric field between the two shells, altering the potential between them. However, this effect is usually negligible unless the external charges are very large and close to the shells.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
581
  • Introductory Physics Homework Help
Replies
23
Views
344
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
889
  • Introductory Physics Homework Help
Replies
15
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
500
  • Introductory Physics Homework Help
Replies
9
Views
1K
Replies
1
Views
143
Back
Top