Electric field acting on the source charge

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The discussion focuses on calculating the force acting on the northern hemisphere of a uniformly charged sphere due to the southern hemisphere. The approach involves using Gauss's law to determine the electric field and integrating the force over the northern hemisphere. It is clarified that while the northern hemisphere generates its own electric field, this internal interaction cancels out, allowing for the calculation to consider only the force exerted by the southern hemisphere. The method involves breaking down the northern hemisphere into smaller fractions to assess the forces acting on each, ultimately leading to the conclusion that the force on the northern hemisphere is solely due to the southern hemisphere. This reasoning aligns with Newton's Third Law, where internal forces within the northern hemisphere negate each other.
Mayan Fung
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I am reading Griffith's textbook on EM. There is a problem asking to find the force acting on the northern hemisphere by the southern hemisphere of a uniformly charged sphere.

The solution idea is to find the expression of the E field by Gauss's law and integrate the force over the northern hemisphere. However, part of the total electric field is contributed by the northern hemisphere itself. In my understanding, we should not include the electric field of the object we are calculating the force on it. I wonder why we can solve this problem with this approach?
 
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Chan Pok Fung said:
In my understanding, we should not include the electric field of the object we are calculating the force on it.
I think you are thinking of point particles.
 
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There is a straight-forward explanation if we break down the process of "calculating the force":

- In order to calculate the force acting on the Northern Hemisphere (NH), we can break the NH down into tiny fractions and calculate the "forces that act on each of those fractions", including:
--> The force that the Southern Hemisphere (SH) act on the fraction
--> The forces between NH fractions

- Then, we can calculate the force that act on the HM by adding all the "forces that act on each of those fractions". In doing so, we will have two sums:
--> Total force that the SH act on all the fractions of NH
===> This is the force that SH act NH as a whole
--> Total force of interaction between NH fractions
===> According to Newton's Third Law of Motion, every force has a counter-force, so the sum of the internal forces of all those fractions would ultimately add up to zero.

In short, we can claim that the force acting on the NH was from the SH (although there are internal interaction in the NH, those interactions canceled each other).
 
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Vinh Nguyen said:
There is a straight-forward explanation if we break down the process of "calculating the force":

- In order to calculate the force acting on the Northern Hemisphere (NH), we can break the NH down into tiny fractions and calculate the "forces that act on each of those fractions", including:
--> The force that the Southern Hemisphere (SH) act on the fraction
--> The forces between NH fractions

- Then, we can calculate the force that act on the HM by adding all the "forces that act on each of those fractions". In doing so, we will have two sums:
--> Total force that the SH act on all the fractions of NH
===> This is the force that SH act NH as a whole
--> Total force of interaction between NH fractions
===> According to Newton's Third Law of Motion, every force has a counter-force, so the sum of the internal forces of all those fractions would ultimately add up to zero.

In short, we can claim that the force acting on the NH was from the SH (although there are internal interaction in the NH, those interactions canceled each other).

That's very clear. Thanks!
 
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