a)A solid sphere, made of an insulating material, has a volume charge density of ? = a/r, where r is the radius from the center of the sphere, a is constant, and a > 0. What is the electric field within the sphere as a function of the radius r? Note: The volume element dV for a spherical shell of radius r and thickness dr is equal to 4?r2dr. (Use the following as necessary: a, r, and ?0.) b.)What If? What if the charge density as a function of r within the charged solid sphere is given by ? = a/r^2? Find the new magnitude and direction of the electric field within the sphere at radius r. (Use the following as necessary: a, r, and ?0.)

Answered: a) A solid sphere, made of an…

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter25: Gauss’s Law

Section: Chapter Questions

Problem 43PQ: The nonuniform charge density of a solid insulating sphere of radius R is given by = cr2 (r R),...

See similar textbooks

Related questions

Concept explainers

Question

A solid sphere, made of an insulating material, has a volume charge density of

? = a/r,

where r is the radius from the center of the sphere, a is constant, and

a > 0.

What is the electric field within the sphere as a function of the radius r? Note: The volume element dV for a spherical shell of radius r and thickness dr is equal to 4?r²dr. (Use the following as necessary: a, r, and ?₀.)

b.)

What If? What if the charge density as a function of r within the charged solid sphere is given by

? = a/r^2?

Find the new magnitude and direction of the electric field within the sphere at radius r. (Use the following as necessary: a, r, and ?₀.)

Expert Solution

Step 1

ρ = volume charge density = a/r

small charge enclosed is given as

dq = ρ dV

dq = (a/r ) ( 4πr²dr )

dq = 4π a r dr

total charge is given as

Q = $\int_{0}^{r}$ 4π a r dr

Q = 4π a r²/2

using gauss's law

E A = Q/ $\epsilon _{o}$

E (4π r²) = 4π a r²/(2 $\epsilon _{o}$ )

E = a/(2 $\epsilon _{o}$ )

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 8 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions