Work Done by an Insulating Sphere on a test charge

AI Thread Summary
The discussion revolves around calculating the work done by the electric field of an insulating sphere on a point charge released from rest outside its surface. The sphere has a uniform charge density of 6.50×10−9 and a radius of 0.240. Participants suggest using the formula kQq/r, treating the sphere as a point charge at its center, and emphasize the importance of calculating the total charge using the volume of the sphere. Clarifications are made regarding the distinction between volume and surface area in this context. The conversation concludes with the original poster expressing gratitude for the assistance received.
tyler3902
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Homework Statement



An insulating sphere of radius 0.240 has uniform charge density 6.50×10−9 . A small object that can be treated as a point charge is released from rest just outside the surface of the sphere. The small object has positive charge 4.10×10−6

How much work does the electric field of the sphere do on the object as the object moves to a point very far from the sphere?

Homework Equations





The Attempt at a Solution



I tried to use kQq/r to find the work done but I'm honestly very lost about where to start.


Any help is extremely appreciated, thanks.
 
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tyler3902 said:

Homework Statement



An insulating sphere of radius 0.240 has uniform charge density 6.50×10−9 . A small object that can be treated as a point charge is released from rest just outside the surface of the sphere. The small object has positive charge 4.10×10−6

How much work does the electric field of the sphere do on the object as the object moves to a point very far from the sphere?

Homework Equations




The Attempt at a Solution



I tried to use kQq/r to find the work done but I'm honestly very lost about where to start.


Any help is extremely appreciated, thanks.

You've got the right formula, now use it. You can treat the sphere as a point charge concentrated at the center of the sphere. What's the total charge of the sphere?
 
would it be the charge density multiplied by the surface area of the sphere?, so 4/3pir^3(6.5x10^-9)?
 
tyler3902 said:
would it be the charge density multiplied by the surface area of the sphere?, so 4/3pir^3(6.5x10^-9)?

4/3pir^3 is the volume of the sphere, not the surface area. But I still think that's the correct thing to do. If it's an insulating sphere they probably gave you the volume density not the surface density.
 
bleh stupid me, appreciate all the help Dick :)

Close please solved
 
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