A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 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
 3 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.

This Question is Open

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

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

anonymous
 3 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.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.