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crisfelicitas
Group Title
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.
 one year ago
 one year ago
crisfelicitas Group Title
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.
 one year ago
 one year ago

This Question is Open

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.1
Use Gauss law to find value of field E(r).
 one year ago

telltoamit Group TitleBest ResponseYou've already chosen the best response.0
You will need to use integral calculus to find charge at distance r
 one year ago

Diwakar Group TitleBest ResponseYou've already chosen the best response.2
a)\[Q=\int\limits_{}^{} \rho dV\] \[dV=4 \pi r ^{2} dr\] 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.
 one year ago
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