What Happens to a Point Charge Inside a Uniformly Charged Sphere?

AI Thread Summary
A point charge placed inside a uniformly charged sphere experiences different behaviors depending on the value of 'a' in the force equation F ∝ 1/r^a. If a > 2, the point charge moves toward the center of the sphere, while if a < 2, it moves toward the surface. This behavior also reverses if the charges have opposite signs. The discussion emphasizes the need for integration of the force over the sphere to derive a formula involving 'a'. Understanding the implications of 'a' is crucial for solving the problem effectively.
danilorj
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Homework Statement



Suppose a strength that acts between two charges depends on the distance, 1/r^a, where
1) a>2
2)a<2
What will happen with a punctual charge, if it is put within a uniformly charged sphere. At the initial moment the punctual charge is in rest.

Homework Equations



F is proportional to 1/r^a.

The Attempt at a Solution


The solution says that for a> 2 the punctual charge q goes toward point O, that it is the center of the sphere, in case of same signals.
and for a<2 the punctual charge goes toward point B that is in the surface of the sphere.
And if the signals are opposite it happens the opposite too.
I don't understand this. Why of going toward the center and the surface of the sphere?
 
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hi danilorj! :smile:

(btw, we don't say a "punctual charge", we say a "point charge" …

"punctual" means not early and not late, for an appointment :wink:)

do an integration, of the force from a small area dθdφ, over the whole sphere …

what do you get? :smile:
 
oh man thanks for the warning.. it is point charge.

But still don't understand what to do with this force. Is it necessary to calculate? The problem states that is proportional to 1/r^a, I don't know about this 'a' whether it can be less than one or even negative.
 
hi danilorj! :smile:

(just got up :zzz:)
danilorj said:
The problem states that is proportional to 1/ra, I don't know about this 'a' whether it can be less than one or even negative.

(try using the X2 button just above the Reply box :wink:)

just leave it as "a", and do the integration …

the result will be a formula using "a" :wink:
 
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