Question on electrostatic pressure

In summary, the conversation is about section 2.5 of Griffiths' electrodynamics, specifically the concept of electrostatic pressure in conductors. A question is posed regarding the force of repulsion between the "northern" and "southern" hemispheres of a metal sphere with a total charge Q. The person speaking has an answer but is unsure of its accuracy and asks for assistance in checking their understanding.
  • #1
brianparks
24
0
I am currently working through section 2.5 of Griffiths' electrodynamics, specifically the part which deals with electrostatic pressure in conductors. I encountered the following question (question 2.39, or 2.38 in earlier versions):

A metal sphere of radius R carries a total charge Q. What is the force of repulsion between the "northern" hemisphere and the "southern" hemisphere?

I have an answer to the question, but I am not sure if it is right. Would anyone be willing to work through the problem, so that I can test my understanding of the subject matter?

Any help is greatly appreciated.

--Brian
 
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  • #2
Post your solution and we'll check it out.
 
Last edited:
  • #3


Sure, I would be glad to help you work through this problem. First, let's review the concept of electrostatic pressure in conductors. In a conductor, the charges are free to move and redistribute themselves until they reach a state of electrostatic equilibrium. This means that the electric field inside the conductor is zero and the charges are distributed uniformly on the surface.

Now, let's apply this concept to the problem at hand. We have a metal sphere of radius R carrying a total charge Q. The charges will redistribute themselves on the surface of the sphere until they reach a state of electrostatic equilibrium. The electric field inside the conductor will be zero, and the charges will be distributed uniformly on the surface.

Now, let's divide the sphere into two halves - the northern hemisphere and the southern hemisphere. Each half will have a charge of Q/2, since the total charge Q is distributed uniformly on the surface of the sphere.

According to Coulomb's law, the force of repulsion between two charges Q1 and Q2 separated by a distance r is given by:

F = (kQ1Q2)/r^2

In this case, Q1 = Q2 = Q/2, and r is the distance between the centers of the two hemispheres. Since the distance between the centers is equal to the radius R, we can rewrite the equation as:

F = (kQ^2)/(4R^2)

Therefore, the force of repulsion between the two hemispheres is given by:

F = (kQ^2)/(4R^2)

I hope this helps you understand the problem better. Let me know if you have any further questions or if you would like me to go through the problem in more detail. Keep up the good work!
 

Related to Question on electrostatic pressure

1. What is electrostatic pressure?

Electrostatic pressure is the force per unit area exerted by electric charges that are at rest in a given space. It is a measure of the repulsion or attraction between these charges and is caused by the electric field they create.

2. How is electrostatic pressure different from other types of pressure?

Unlike other types of pressure, such as gas or liquid pressure, which are caused by the motion of particles, electrostatic pressure is caused by the static position of electric charges. It is also a vector quantity, meaning it has both magnitude and direction.

3. What factors affect the magnitude of electrostatic pressure?

The magnitude of electrostatic pressure depends on the amount of charge present, the distance between the charges, and the permittivity of the medium between the charges. It also varies with the geometry of the charges and their distribution.

4. How is electrostatic pressure calculated?

Electrostatic pressure can be calculated using the formula P = kQ2/Aε0d2, where P is the electrostatic pressure, k is the Coulomb's constant, Q is the total charge, A is the area of the surface, ε0 is the permittivity of free space, and d is the distance between the charges.

5. What are some real-world applications of electrostatic pressure?

Electrostatic pressure has many practical applications, including electrostatic precipitators used for air pollution control, inkjet printers, and electrostatic motors. It is also important in understanding the behavior of plasma in fusion reactors and in the design of high-voltage power lines.

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