Chapter 16, Problem 89QAP

To determine

(a)

An expression for the electric field in the radial region ri.

Answer to Problem 89QAP

An expression for the electric field in the radial region ri is, $E_{r<R_i}=0$

Explanation of Solution

Given:

Radius of inner sphere, $R_i$

Surface Charge density for inner sphere, $+{\sigma }_i$

Radius of outer sphere, $R_o$

Surface Charge density for outer sphere, $-{\sigma }_o$

Formula used:

The electric force is given by,
  $E=\frac{\sigma }{2{\epsilon }_0}$

Where,
  $E$ =Electric field
  $\sigma$ = Surface charges density
  ${\epsilon }_0$ =Permittivity of free space

Calculation:

The sphere is conducting. The charge enclosed by the sphere is zero. Thus, the electric field inside the sphere is zero.
  $E_{r<R_i}=0$

To determine

(b)

An expression for the electric field in the radial region Rio.

Answer to Problem 89QAP

An expression for the electric field in the radial region Rio is, $E_{Ri<r<R_o}=\frac{{\sigma }_i{R}_i^2}{{\epsilon }_0{r}^2}$ and directed outside from the sphere.

Explanation of Solution

Given:

Radius of inner sphere, $R_i$

Surface Charge density for inner sphere, $+{\sigma }_i$

Radius of outer sphere, $R_o$

Surface Charge density for outer sphere, $-{\sigma }_o$

Formula used:

The electric force is given by,
  $E=\frac{Q}{4\pi {\epsilon }_0{r}^2}$

Where,
  $E$ =Electric field
  $Q$ = Charge
  ${\epsilon }_0$ =Permittivity of free space
r=Distance

Calculation:

The electric force is given by,
  $E=\frac{Q}{4\pi {\epsilon }_0{r}^2}$

But, $Q={\sigma }_{i}A$

  $\begin{array}{l}\therefore E=\frac{{\sigma }_{i}4\pi R_i^2}{4\pi {\epsilon }_0{r}^2}\\ E_{Ri<r<R_o}=\frac{{\sigma }_i{R}_i^2}{{\epsilon }_0{r}^2}\end{array}$

It is directed outside from the sphere.

To determine

(c)

An expression for the electric field in the radial region r>R_o.

Answer to Problem 89QAP

An expression for the electric field in the radial region r>R_ois, $E=\frac{{\sigma }_i{R}_i^2-{\sigma }_o{R}_o^2}{{\epsilon }_0{r}^2}$

Explanation of Solution

Given:

Radius of inner sphere, $R_i$

Surface Charge density for inner sphere, $+{\sigma }_i$

Radius of outer sphere, $R_o$

Surface Charge density for outer sphere, $-{\sigma }_o$

Formula used:

The electric force is given by,
  $E=\frac{Q}{4\pi {\epsilon }_0{r}^2}$

Where,
  $E$ =Electric field
  $Q$ = Charge
  ${\epsilon }_0$ =Permittivity of free space
r=DistanceCalculation:

The electric field is given by,
  $E=\frac{Q}{4\pi {\epsilon }_0{r}^2}$

But, $Q={\sigma }_{i}A$

  $\begin{array}{l}\therefore E_i=\frac{{\sigma }_{i}4\pi R_i^2}{4\pi {\epsilon }_0{r}^2}\\ E_i=\frac{{\sigma }_i{R}_i^2}{{\epsilon }_0{r}^2}\end{array}$

And
  $\begin{array}{l}\therefore E_o=\frac{{\sigma }_{i}4\pi R_o^2}{4\pi {\epsilon }_0{r}^2}\\ E_o=\frac{{\sigma }_i{R}_o^2}{{\epsilon }_0{r}^2}\end{array}$

The electric field is given by,
  $\begin{array}{l}E=E_i-E_o\\ E=\frac{{\sigma }_i{R}_i^2}{{\epsilon }_0{r}^2}-\frac{{\sigma }_o{R}_o^2}{{\epsilon }_0{r}^2}\\ E=\frac{{\sigma }_i{R}_i^2-{\sigma }_o{R}_o^2}{{\epsilon }_0{r}^2}\end{array}$

If the value of ${\sigma }_i{R}_i^2-{\sigma }_o{R}_o^2>1$ The field is directed outwards.

If the value of ${\sigma }_i{R}_i^2-{\sigma }_o{R}_o^2<1$ The field is directed inwards.