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anonymous
 4 years ago
An insulating sphere of radius R has spherically symmetric but radially nonuniform distribution of charge, with a charge density ρ(r) given by
ρ(r)=ρ0[1−(5/4)(r/R)^2].
The constant ρ0 > 0 is the charge density at the center of the sphere. Note that the charge density ρ(r) is negative for 2/√5R < r <R. In the following, express your results analytically in terms of any or all of ε0, ρ0, R and r.
a) What is the total charge contained in the sphere? Is the total charge positive or negative?
b) Write expressions for the magnitude of the electric field as a function of r for r < R.
anonymous
 4 years ago
An insulating sphere of radius R has spherically symmetric but radially nonuniform distribution of charge, with a charge density ρ(r) given by ρ(r)=ρ0[1−(5/4)(r/R)^2]. The constant ρ0 > 0 is the charge density at the center of the sphere. Note that the charge density ρ(r) is negative for 2/√5R < r <R. In the following, express your results analytically in terms of any or all of ε0, ρ0, R and r. a) What is the total charge contained in the sphere? Is the total charge positive or negative? b) Write expressions for the magnitude of the electric field as a function of r for r < R.

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VincentLyon.Fr
 4 years ago
Best ResponseYou've already chosen the best response.1Use Gauss law to find value of field E(r).

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You will need to use integral calculus to find charge at distance r

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0a)\[Q=\int\limits_{}^{} \rho dV\] \[dV=4 \pi r ^{2} dr\] Integrate from r=0 to r=R \[Q=4 \pi \rho 0 \int\limits_{0}^{R} (15/4(r/R)^{2})r ^{2}dr\] \[Q=4 \pi \rho0 R ^{3}/12\] Charge is positive b) As told by VincentLyon, we use Gauss thm. here A Gaussian surface is a sphere of radius r and by spherical symmetry we take the value of field to be same at all points on the sphere and RADIALLY outwards. \[E \times 4\pi r ^{2} = \rho o(r ^{3}/3r ^{5}/4R ^{2})/ \epsilon\] I found the charge inside the sphere the same way as in part a) Solve for E from here. PLEASE TELL ME IF I AM WRONG SOMEWHERE.
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