Consider a uniformly charged sphere with radius R and charge per unit volume . (
ID: 1526501 • Letter: C
Question
Consider a uniformly charged sphere with radius R and charge per unit volume .
(a) What is the magnitude of the electric field a distance
r = R/6
from the center of the sphere? (Use the following as necessary: 0, , R.)
E =
(b) Explain in words why the portion of the sphere that is more than
r = R/6
from the sphere's center does not contribute to the net electric field at
r = R/6.
a) The outer portion does not contribute to the net electric field at r because the flux only not passes into and not out of the Gaussian surface.
b) The outer portion does not contribute to the net electric field at r because the flux of the outer portion passes into and out of the Gaussian surface.
3)The outer portion does not contribute to the net electric field at r because the flux of the outer portion only points outward and not toward the inner portion.
(c) Now suppose a point charge is placed a distance
r = 2R
from the center of the sphere.
Will this point charge change the electric field in part (a)? Explain why or why not.
Yes, the electric field from the new charge is no longer radial and therefore passes unevenly through the Gaussian surface from part (a).No, since the new charge is outside of the Gaussian surface from part (a) nothing would change.
Explanation / Answer
a) According to Gauss Law
E.A = Q/0
=> E * (4pi * r2) = ( * 4/3pi * r3)/0
=> Electric field , E = ( * 1/3 * r)/0
Thus, the magnitude of the electric field a distance r = R/6 = ( * R)/180
b) Correct answer is : The outer portion does not contribute to the net electric field at r because the flux of the outer portion only points outward and not toward the inner portion.
c) Correct Answer is : Yes, the electric field from the new charge is no longer radial and therefore passes unevenly through the Gaussian surface from part (a).