Electric Field and Potential due to Induced Charges

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An uncharged metal spherical body will have induced charges on its surface when a charge q is placed outside it, resulting in an electric field inside the body that is zero. The electric field due to the external charge q is given by E = Kq/r^2, and the induced charges create an opposing electric field of equal magnitude. The discussion raises the question of whether the electric field due to induced charges is from the positive or negative induced charge. It is noted that the induced charge is distributed unevenly across the surface, complicating the calculation of the electric field from either the positive or negative induced charges alone. Understanding the behavior of induced charges is essential for solving problems related to electric fields in electrostatics.
gracy
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Homework Statement


If there is an uncharged metal spherical body and a charge q outside it .Find Electric field due to induced charges.

Homework Equations


##E##=##\frac{Kq}{r^2}##

The Attempt at a Solution


If there is an uncharged metal spherical body and a charge q outside it then there will be induced charges on it's surface like this
UNCHARGED.png

Electric field inside the body should be zero.Electric field due to +q would be ##E##=##\frac{Kq}{r^2}## there fore Electric field due to induced charges should be equal in magnitude to that of ##E##=##\frac{Kq}{r^2}## and opposite in direction.But my question is this electric field due to induced charges is due to +q induced or -q induced?
 
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gracy said:

Homework Statement


If there is an uncharged metal spherical body and a charge q outside it .Find Electric field due to induced charges.

Homework Equations


##E##=##\frac{Kq}{r^2}##

The Attempt at a Solution


If there is an uncharged metal spherical body and a charge q outside it then there will be induced charges on it's surface like this
View attachment 93590
Electric field inside the body should be zero.Electric field due to +q would be ##E##=##\frac{Kq}{r^2}## there fore Electric field due to induced charges should be equal in magnitude to that of ##E##=##\frac{Kq}{r^2}## and opposite in direction.But my question is this electric field due to induced charges is due to +q induced or -q induced?
What does the question ask you to do? "Find Electric field due to induced charges." That must include all of the charges mentioned with the exception of the given charge, +q.
 
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SammyS said:
That must include all of the charges mentioned with the exception of the given charge, +q.
And what if we are asked to find electric field due to +q induced?
 
gracy said:
And what if we are ask to find electric field due to +q induced?
The induced charge is distributed over the surface with varying density. It would be a very involved process to find the electric field due to +qinduced by itself or the electric field due to -qinduced by itself.
 
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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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