A solid insulating sphere of radius R has a non-uniform charge density ρ = Ar2 , where A is a constant and r is measured from the center of the sphere.
a) Show that the electric field outside the sphere (r > R) is E = AR5 /(5εor 2 ).
b) Show that the electric field inside the sphere (r < R) is E = AR3 /(5εo).
Hint: The total charge Q on the sphere is found by integrating ρ dV, over appropriate limits inside or outside the sphere. Also, dV can be treated as very thin, spherical shells where dV = 4πr 2 dr.
Please be as clear and detailed as possible, explaining each step
for second part consider a
Gausian surface which is a sphere of radius r , concentric with
given charged sphere. Then Gauss' Law is applied.
We have to take the charge enclosed within the Gaussian surface ( from r=0 to r = r)
A solid insulating sphere of radius R has a non-uniform charge density ρ = Ar2 ,...
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3rd Question
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Consider a charged sphere with the following charge density ρ(r)
=(ρ0(1− r Rmax) r ≤ Rmax 0 r > Rmax
Using Gauß’ law, calculate the electric field
(a) ~ E1 inside the sphere (i.e. r ≤ Rmax),
(b) ~ E2 outside the sphere (i.e r ≥ Rmax),
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Please provide an explanation for the
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(radius R) with non-uniform charge density rho=beta*r^2 with
beta>0.
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