Chapter 26, Problem 59PQ

(a)

To determine

The magnitude of the electric field and electric potential.

Answer to Problem 59PQ

The magnitude of the electric field is $0$ and electric potential is $3.45\times {10}^4\text{V}$.

Explanation of Solution

The electric field inside a hollow conductor is zero.

Write an expression for the electric potential.

  $V=k\frac{Q}{R}$                                                                                                              (I)

Here, $k$ is the culoumb constant, $Q$ is charge, $R$ is the radius of the sphere and $V$ is the electric potential.

Conclusion:

The radius of the conductor is $22.0\text{cm}$ and the radial distance is $5.00\text{cm}$. Thus, the point is inside the conductor. Thus, the electric field at $5.00\text{cm}$ is zero.

Sunstitute $8.99\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $845\text{nC}$ for $Q$ and $22.0\text{cm}$ for $R$ in equation (I) to find $V$.

    $\begin{array}{c}V=\left(8.99\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(\left(845\text{nC}\right)\left(\frac{1\text{C}}{{10}^9\text{nC}}\right)\right)}{\left(\left(22.0\text{cm}\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)}\\ =\left(8.99\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(845\times {10}^{-9}\text{C}\right)}{\left(22.0\times {10}^{-2}\text{m}\right)}\\ =3.45\times {10}^4\text{V}\end{array}$

Thus, the magnitude of the electric field is $0$ and electric potential is $3.45\times {10}^4\text{V}$.

(b)

To determine

The magnitude of the electric field and electric potential.

Answer to Problem 59PQ

The magnitude of the electric field is $1.57\times {10}^5\text{V/m}$ and electric potential is $3.45\times {10}^4\text{V}$.

Explanation of Solution

Write an expression for the electric field.

  $E=k\frac{Q}{r^2}$                                                                                                             (II)

Here, $r$ is the radial distance and $E$ is the electric potential.

Write an expression for the electric potential.

  $V=k\frac{Q}{r}$                                                                                                            (III)

Conclusion:

Sunstitute $8.99\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $845\text{nC}$ for $Q$ and $22.0\text{cm}$ for $R$ in equation (II) to find $E$.

    $\begin{array}{c}E=\left(8.99\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(\left(845\text{nC}\right)\left(\frac{1\text{C}}{{10}^9\text{nC}}\right)\right)}{{\left(\left(22.0\text{cm}\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)}^2}\\ =\left(8.99\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(845\times {10}^{-9}\text{C}\right)}{{\left(22.0\times {10}^{-2}\text{m}\right)}^2}\\ =1.57\times {10}^5\text{V/m}\end{array}$

Sunstitute $8.99\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $845\text{nC}$ for $Q$ and $22.0\text{cm}$ for $R$ in equation (III) to find $V$.

    $\begin{array}{c}V=\left(8.99\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(\left(845\text{nC}\right)\left(\frac{1\text{C}}{{10}^9\text{nC}}\right)\right)}{\left(\left(22.0\text{cm}\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)}\\ =\left(8.99\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(845\times {10}^{-9}\text{C}\right)}{\left(22.0\times {10}^{-2}\text{m}\right)}\\ =3.45\times {10}^4\text{V}\end{array}$

Thus, the magnitude of the electric field is $1.57\times {10}^5\text{V/m}$ and electric potential is $3.45\times {10}^4\text{V}$.

(c)

To determine

The magnitude of the electric field and electric potential.

Answer to Problem 59PQ

The magnitude of the electric field is $3.04\times {10}^4\text{V/m}$ and electric potential is $1.52\times {10}^4\text{V}$.

Explanation of Solution

Write an expression for the electric field.

  $E=k\frac{Q}{r^2}$                                                                                                           (IV)

Write an expression for the electric potential.

  $V=k\frac{Q}{r}$                                                                                                            (V)

Conclusion:

Sunstitute $8.99\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $845\text{nC}$ for $Q$ and $50.0\text{cm}$ for $R$ in equation (IV) to find $E$.

    $\begin{array}{c}E=\left(8.99\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(\left(845\text{nC}\right)\left(\frac{1\text{C}}{{10}^9\text{nC}}\right)\right)}{{\left(\left(50.0\text{cm}\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)}^2}\\ =\left(8.99\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(845\times {10}^{-9}\text{C}\right)}{{\left(50.0\times {10}^{-2}\text{m}\right)}^2}\\ =3.04\times {10}^4\text{V/m}\end{array}$

Sunstitute $8.99\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $845\text{nC}$ for $Q$ and $50.0\text{cm}$ for $R$ in equation (V) to find $V$.

    $\begin{array}{c}V=\left(8.99\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(\left(845\text{nC}\right)\left(\frac{1\text{C}}{{10}^9\text{nC}}\right)\right)}{\left(\left(50.0\text{cm}\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)}\\ =\left(8.99\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(845\times {10}^{-9}\text{C}\right)}{\left(50.0\times {10}^{-2}\text{m}\right)}\\ =1.52\times {10}^4\text{V}\end{array}$

Thus, the magnitude of the electric field is $3.04\times {10}^4\text{V/m}$ and electric potential is $1.52\times {10}^4\text{V}$.