Solution Found!
CALC An insulating hollow sphere has inner radius a and
Chapter 22, Problem 53P(choose chapter or problem)
CALC An insulating hollow sphere has inner radius \(a\) and outer radius \(b\). Within the insulating material the volume charge density is given by \(\rho(r)=\frac{\alpha}{r}\), where \(\alpha\) is a positive constant.
(a) In terms of \(\alpha\) and \(a\), what is the magnitude of the electric field at a distance \(r\) from the center of the shell, where \(a<r<b\)?
(b) A point charge \(q\) is placed at the center of the hollow space, at \(r=0\). In terms of \(\alpha\) and \(a\), what value must \(q\) have (sign and magnitude) in order for the electric field to be constant in the region \(a<r<b\), and what then is the value of the constant field in this region?
Questions & Answers
QUESTION:
CALC An insulating hollow sphere has inner radius \(a\) and outer radius \(b\). Within the insulating material the volume charge density is given by \(\rho(r)=\frac{\alpha}{r}\), where \(\alpha\) is a positive constant.
(a) In terms of \(\alpha\) and \(a\), what is the magnitude of the electric field at a distance \(r\) from the center of the shell, where \(a<r<b\)?
(b) A point charge \(q\) is placed at the center of the hollow space, at \(r=0\). In terms of \(\alpha\) and \(a\), what value must \(q\) have (sign and magnitude) in order for the electric field to be constant in the region \(a<r<b\), and what then is the value of the constant field in this region?
ANSWER:Solution 53P Step 1 of 3: First apply the Gauss’s law and take a spherical Gaussian surface because of the spherical symmetry of the charge distribution. The net field is the vector sum of the field due to q and the field due to the sphere. r 2 (r) = , rv = 4r dr and Q = ( )dv a