Work done enclosing sphere with spherical shell - for tommorow

It's always a great feeling when you finally figure out a difficult problem. Congrats on cracking it!In summary, Gareth asks for help with a problem on Electromagnetism and asks for pointers on how to get started. He suggests integrating over a spherical shell and then integrating from r to a. Another user offers a hint on finding the work done in bringing the spherical shell and the force on it. Gareth later returns to share that he has solved the problem and expresses his joy in solving it.
  • #1
ghosts_cloak
16
0
Hi guys!
I have been working on this question all day and am getting no where :cry: I really can't get to grips with Electromagnetism, arghh!
The question is :http://www.zen96175.zen.co.uk/problem.GIF
I would NOT like anyone to post a solution as its assessed work but I would IMMENSELY appreciate some pointers on how to get started on this...
I have a few hours left to complete this, I hope somebody has a few pointers :!) :smile:
Thank you ,

~Gaz
 
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  • #2
So what have you done with this problem? If you show where you are stuck, it will be easier for us to help.
 
  • #3
Hiya,
Well, that's the point really, I need a little push to get me started!
I am thinking it involves seeing how much work it takes to bring a hoop of charge from the spherical shell to the sphere, and then integrating over the whole shell. And then, integrate from r to a to make up the final large sphere is mentions? Pretty vague I know :cry:
Just looking for some pointers if anyone can!

Thank you!
~Gareth
 
  • #4
ghosts_cloak said:
Hiya,
Well, that's the point really, I need a little push to get me started!
I am thinking it involves seeing how much work it takes to bring a hoop of charge from the spherical shell to the sphere, and then integrating over the whole shell. And then, integrate from r to a to make up the final large sphere is mentions? Pretty vague I know :cry:
Just looking for some pointers if anyone can!

Thank you!
~Gareth

I don't think you need to bring a hoop of charge, but you are on the right track. First find the work done in bringing the spherical shell from [itex] \infty [/itex] to [itex] r [/itex] and then integrate r from 0 to a.

To find the work done in bringing the shell, you will have to know the electric field due to sphere at the center, at a distance x and also the charge dq on the spherical shell (Hint: when the shell has finally enveloped the sphere, the charge dq is still going to be the same, so you can write dq in terms of r and dr). Then you can find the force on the spherical shell and hence the work done in bringing it to the sphere

I have not evaluated the double integral myself, but I think it should give you the answer.
 
Last edited:
  • #5
Phew, after a whole day on it, I think I have cracked it! This feeling is what makes physics and maths the greatest!

Thanks for your help,
~Gareth
 
  • #6
Way to go, Gareth!
 

FAQ: Work done enclosing sphere with spherical shell - for tommorow

What is the equation for calculating the work done enclosing a sphere with a spherical shell?

The equation for calculating the work done enclosing a sphere with a spherical shell is W = 4πε0 (R2 - R1)Q, where W is the work done, ε0 is the permittivity of free space, R2 is the radius of the outer spherical shell, R1 is the radius of the inner sphere, and Q is the charge enclosed in the sphere.

What is the significance of enclosing a sphere with a spherical shell in terms of work done?

Enclosing a sphere with a spherical shell means that the electric field inside the sphere is zero. This results in the work done to be equal to the work done on the charge, which simplifies the calculation and makes it easier to determine the amount of work done.

How does the distance between the inner sphere and outer shell affect the work done?

The distance between the inner sphere and outer shell, (R2 - R1), is directly proportional to the work done, W. This means that as the distance increases, the work done also increases. This relationship is represented by the equation, W = 4πε0 (R2 - R1)Q.

What is the unit of measurement for work done enclosing a sphere with a spherical shell?

The unit of measurement for work done is joules (J). This is because work is defined as the product of force and distance, and both force and distance are measured in units of newtons (N) and meters (m), respectively, which results in the unit of joules for work.

Is the work done enclosing a sphere with a spherical shell always positive?

No, the work done enclosing a sphere with a spherical shell can be positive, negative, or zero. It depends on the direction of the force and displacement of the charge. If the force and displacement are in the same direction, the work done is positive. If they are in opposite directions, the work done is negative. And if there is no displacement, the work done is zero.

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