Chapter 33, Problem 25A

Shown below are three hollow copper spheres. Sphere A has a radius R, Sphere B has a radius of 2R, and Sphere C has a radius 3R. On each sphere is a charge, as indicated, which is evenly distributed over the spheres surface. (Each sphere is independent of the others; they don’t influence one another.)

a. Rank from greatest to least the magnitude of the electric fields at a distance 4R from the center of the spheres.

b. Rank the field strengths at the center of the spheres.

c. Rank the potentials at distance 4R (assuming the potential at infinity is zero).

d. Suppose the charge is redistributed so that all three spheres have identical charges. Rank the fields at distance 4R from greatest to least.

To determine

To find: The rank of each sphere on the basis of electric field at 4R.

Explanation of Solution

The Electric field of sphere, $E=\frac{kq}{r^2}$ , where $r$ is the distance from the center of sphere.

Electric field of sphere is as follow,

  $\begin{array}{l}{E}_A=\frac{k\left(2q\right)}{{\left(4R\right)}^2}=0.125\frac{kq}{R^2}\\ E_B=\frac{k\left(q\right)}{{\left(4R\right)}^2}=0.0625\frac{kq}{R^2}\\ E_C=\frac{k\left(3q\right)}{{\left(4R\right)}^2}=0.1875\frac{kq}{R^2}\end{array}$

The rank of each sphere on the basis of electric field at 4R is

  $E_C>E_A>E_B$

To determine

To find: The rank of each sphere on the basis of electric field at center.

Explanation of Solution

Electric field inside hollow sphere is zero

Therefore, all the sphere electric field is zero

  $E_A=E_B=E_c=0$

To determine

To find: The rank of each sphere on the basis of potential at 4R.

Explanation of Solution

The voltage of charged sphere, $V=\frac{kq}{r}$ , where $r$ is the distance from the center of sphere.

Voltage of sphere is as follow,

  $\begin{array}{l}{V}_A=\frac{k\left(2q\right)}{\left(4R\right)}=0.5\frac{kq}{R}\\ V_B=\frac{k\left(q\right)}{\left(4R\right)}=0.25\frac{kq}{R}\\ V_C=\frac{k\left(3q\right)}{\left(4R\right)}=0.75\frac{kq}{R}\end{array}$

The rank of each sphere on the basis of voltage at 4R is

  $V_C>V_A>V_B$

To determine

To find: The rank of each sphere on the basis of potential at 4R if all sphere have same charge.

Explanation of Solution

The voltage of charged sphere, $V=\frac{kq}{r}$ , where $r$ is the distance from the center of sphere.

Voltage of sphere is as follow,

  $\begin{array}{l}{V}_A=\frac{k\left(q\right)}{\left(4R\right)}=0.25\frac{kq}{R}\\ V_B=\frac{k\left(q\right)}{\left(4R\right)}=0.25\frac{kq}{R}\\ V_C=\frac{k\left(q\right)}{\left(4R\right)}=0.25\frac{kq}{R}\end{array}$

All the sphere have same potential at 4R.

  $V_C=V_A=V_B$