Using Gauss\' Law, determine the electric field at various points in the diagram

Question

Using Gauss' Law, determine the electric field at various points in the diagram below, which illustrates a point charge at the center of an initially uncharged conducting sphere. Start by labeling the radii of each point (r_0, r_1, r_6). Then, consider a Gaussian Surface around the center with radius given as the radius to that point. As radius increases what happens to the surface area of the Gaussian Surface? Consequently, what happens to the charge enclosed by the surface? Then use Gauss' Law to determine the electric field strength. You will also need to consider how the charge within the conductor gets distributed.

Explanation / Answer

surface area = 4 pi r^2

as radius increases, Area also increases. ...........Ans

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charge enclosed will remain same. ......Ans

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from Gauss law,

E. A = Qin / e0

E ( 4pi r^2) = q / e0

E = q / (4 pi e0 r^2)

for r1:

E = q / (4 pi e0 r1^2 )

for r2:

E = q / (4 pi e0 r2^2 )

for r3:

E = 0 (inside the conductor)

for r4:

E = 0

for r5:

E = q / (4 pi e0 r5^2)

for r6:

E = q / (4 pi e0 r6^2)

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charge will reside on surafce of conductor.

charge at inside surface, Qin = - q

charge at outside surface, Qout = + q

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