Chapter 20, Problem 32P

(a)

To determine

The electric field and electric potential at $r=10.00\text{cm}$.

Answer to Problem 32P

The electric field at $r=10.00\text{cm}$ is zero and the electric potential is $1.67\text{MV}$.

Explanation of Solution

Given information: The radius of the spherical conductor is $14.00\text{cm}$ and charge is $26\text{μC}$.

In any conducting body like cylinder and shell the charge exist only on the surface of the body. At any point within the sphere, the charge is zero. Charge exists only on the surface of the sphere.

Therefore, at $r=10.00\text{cm}$ the electric field is zero.

Write the expression to calculate the electric potential.

$V=\frac{KQ}{r}$ (1)

Here,

$K$ is the Coulomb’s law constant.

$r$ is the separation between the charges.

$Q$ is the charge.

$V$ is the electric potential.

Substitute $9\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $K$, $26\text{μC}$ for $Q$ and $14.00\text{cm}$ for $r$ in equation (1) to find $V$.

$\begin{array}{c}V=\frac{9\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\left(26.00\text{μC}\times \frac{{10}^{-6}\text{C}}{1\text{μC}}\right)}{\left(14.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}\\ =1.67\text{MV}\end{array}$

Thus, the electric potential is $1.67\text{MV}$.

Conclusion:

Therefore, the electric field at $r=10.00\text{cm}$ is zero and the electric potential is $1.67\text{MV}$.

(b)

To determine

The electric field and electric potential at $r=20.00\text{cm}$.

Answer to Problem 32P

The electric field at $r=20.00\text{cm}$ is $5.84\text{M}\text{N}/\text{C}$ and the electric potential is $1.67\times {10}^6\text{V}$.

Explanation of Solution

Given information: The radius of the spherical conductor is $14.00\text{cm}$ and charge is $26\text{μC}$.

Write the expression to calculate the electric field intensity.

$E=\frac{KQ}{r^2}$ (2)

Here,

$E$ is the electric field intensity.

$K$ is the Coulomb’s law constant.

$r$ is the separation between the charges.

$Q$ is the charge.

Substitute $9\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $K$, $26\text{μC}$ for $Q$ and $20.00\text{cm}$ for $r$ in equation (2).

$\begin{array}{c}E=\frac{9\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\left(26.00\text{μC}\times \frac{{10}^{-6}\text{C}}{1\text{μC}}\right)}{{\left(20.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2}\\ =5.84\text{MN}/\text{C}\end{array}$

Thus, the electric field at $r=20.00\text{cm}$ is $5.84\text{MN}/\text{C}$.

Write the expression to calculate the electric potential.

$V=\frac{KQ}{r}$

Substitute $9\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $K$, $26\text{μC}$ for $Q$ and $20.00\text{cm}$ for $r$ in above equation to find $V$.

$\begin{array}{c}V=\frac{9\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\left(26.00\text{μC}\times \frac{{10}^{-6}\text{C}}{1\text{μC}}\right)}{\left(20.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}\\ =1.17\text{MV}\end{array}$

Thus, the electric potential is $1.17\text{MV}$.

Conclusion:

Therefore, the electric field at $r=20.00\text{cm}$ is $5.84\text{MN}/\text{C}$ and the electric potential is $1.17\text{MV}$.

(c)

To determine

The electric field and electric potential at $r=14.00\text{cm}$.

Answer to Problem 32P

The electric field at $r=14.00\text{cm}$ is $11.9\text{MN}/\text{C}$ and the electric potential is $1.67\text{MV}$.

Explanation of Solution

Given information: The radius of the spherical conductor is $14.00\text{cm}$ and charge is $26\text{μC}$.

Write the expression to calculate the electric field intensity.

$E=\frac{KQ}{r^2}$

Here,

$E$ is the electric field intensity.

$K$ is the Coulomb’s law constant.

$r$ is the separation between the charges.

$Q$ is the charge.

Substitute $9\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $K$, $26\text{μC}$ for $Q$ and $14.00\text{cm}$ for $r$ in equation (2).

$\begin{array}{c}E=\frac{9\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\left(26.00\text{μC}\times \frac{{10}^{-6}\text{C}}{1\text{μC}}\right)}{{\left(14.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2}\\ =11.9\text{MN}/\text{C}\end{array}$

Thus, the electric field at $r=14.00\text{cm}$ is $11.9\text{MN}/\text{C}$.

Write the expression to calculate the electric potential.

$V=\frac{KQ}{r}$

Substitute $9\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $K$, $26\text{μC}$ for $Q$ and $14.00\text{cm}$ for $r$ in above equation to find $V$.

$\begin{array}{c}V=\frac{9\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\left(26.00\text{μC}\times \frac{{10}^{-6}\text{C}}{1\text{μC}}\right)}{\left(14.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}\\ =1.67\text{MV}\end{array}$

Thus, the electric potential is $1.67\text{MV}$.

Conclusion:

Therefore, the electric field at $r=14.00\text{cm}$ is $11.9\text{MN}/\text{C}$ and the electric potential is $1.67\text{MV}$.