Force on a sphere in a constant external electric field

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An uncharged solid sphere placed in a constant external electric field E0 experiences induced charge separation, resulting in the formation of a dipole. The force on the sphere can be calculated using the induced charge and the electric field near the sphere. The equation F = q*E is applicable, where q represents the induced charge and E is the electric field acting on the sphere. The discussion highlights the importance of recognizing that while the external field is uniform, the induced dipole creates a net force on the sphere. Understanding these concepts is crucial for solving the problem effectively.
captain.joco
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Homework Statement


An uncharged solid sphere is paced in a constant external electric field E0. What is the force on the sphere?


Homework Equations





The Attempt at a Solution


I know ( already have found ) the electric potential and electric field around the sphere. Also worked out the surface charged density. I don't know how to calculate the force..Any help please?
 
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Isn't F = q*E ?
 
It seemed too simple somehow... Is q the induced charge on the sphere, and E the electric field close to the sphere??

Thank you for your help
 
captain.joco said:
It seemed too simple somehow... Is q the induced charge on the sphere, and E the electric field close to the sphere??

Thank you for your help

If there is charge separation, what will the net force be?
 
Got it! Thanks a lot for the help!
 
This is the exact question i was looking for help with and still not sure of what the next step is.

I know the uniform E-field causes the sphere to gain an induced dipole. From this how do you get the force? Is it ok to just assume the E-field remains uniform.. this can't be right as there would then be no force and seems there must be a force from the way the question is worded.
 
Thread 'Correct statement about size of wire to produce larger extension'

Thread 'Correct statement about size of wire to produce larger extension'

The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...
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