Chapter 16, Problem 54E

(a)

To determine

The charges on the plates of the capacitor.

Answer to Problem 54E

The charges on the plates is $\underset{\_}{1.0\times {10}^{-4}\text{C}}$.

Explanation of Solution

Write the expression for the capacitance.

  $Q=CV$        (I)

Here, $Q$ is the charge, $C$ is the capacitance, and $V$ is the potential difference.

Conclusion:

Substitute $2\mu \text{F}$ for $C$ and $50\text{V}$ for $V$ in equation (I) to find $Q$.

  $\begin{array}{c}Q=\left(2\mu \text{F}\times \frac{1\text{F}}{1\times {10}^6\mu \text{F}}\right)\left(50\text{V}\right)\\ =1.0\times {10}^{-4}\text{C}\end{array}$

Therefore, the charges on the plates is $\underset{\_}{1.0\times {10}^{-4}\text{C}}$.

(b)

To determine

The electric field between the plates of the capacitor.

Answer to Problem 54E

The electric field between the plates is $\underset{\_}{50000\text{V}\cdot {\text{m}}^{-1}}$.

Explanation of Solution

Write the expression for the electric field is,

  $E=\frac{V}{l}$        (II)

Here, $V$ is the potential difference, $E$ is the electric field, and $l$ is the distance between the plates of the capacitor.

Conclusion:

Substitute $50\text{V}$ for $V$ and ${10}^{-3}\text{m}$ for $l$ in equation (I) to find $V$.

  $\begin{array}{c}E=\frac{50\text{V}}{{10}^{-3}\text{m}}\\ =50000\text{V}\cdot {\text{m}}^{-1}\end{array}$

Therefore, the electric field between the plates is $\underset{\_}{50000\text{V}\cdot {\text{m}}^{-1}}$.

(c)

To determine

The new charge when dielectric constant inserted between the plates.

Answer to Problem 54E

The new charge when dielectric constant inserted between the plates is $\underset{\_}{2.0\times {10}^{-4}\text{C}}$.

Explanation of Solution

Write the expression for the new charge when dielectric constant inserted between the plates is,

  $Q\text{'}=KCV$        (III)

Here, $Q\text{'}$ is the new charge when dielectric constant inserted between the plates of the capacitor, and $K$ is the dielectric field.

Conclusion:

Substitute $2\mu \text{F}$ for $C$, $5$ for $K$, and $50\text{V}$ for $V$ in equation (I) to find $Q$.

  $\begin{array}{c}Q\text{'}=\left(5\right)\left(2\mu \text{F}\times \frac{1\text{F}}{1\times {10}^6\mu \text{F}}\right)\left(50\text{V}\right)\\ =2.0\times {10}^{-4}\text{C}\end{array}$

Therefore, the new charge when dielectric constant inserted between the plates is $\underset{\_}{2.0\times {10}^{-4}\text{C}}$.

(d)

To determine

The electric field between the plates of the capacitor on insertion of dielectric.

Answer to Problem 54E

The electric field between the plates of the capacitor on insertion of dielectric is $\underset{\_}{1.0\times {10}^5\text{V}\cdot {\text{m}}^{-1}}$.

Explanation of Solution

Write the expression for the new potential difference when dielectric constant inserted between the plates is,

  $V\text{'}=\frac{Q\text{'}}{C}$        (IV)

Here, $V\text{'}$ is the new potential difference when dielectric constant inserted between the plates.

Write the expression for the electric field between the plates of the capacitor on insertion of dielectric is,

  $E\text{'}=\frac{V\text{'}}{l}$        (V)

Here, $E\text{'}$ is the electric field between the plates of the capacitor on insertion of dielectric.

Substitute equation (IV) in (V) to get $E\text{'}$,

  $E\text{'}=\frac{Q\text{'}}{Cl}$        (VI)

Conclusion:

Substitute $2.0\times {10}^{-4}\text{C}$ for $Q\text{'}$, ${10}^{-3}\text{m}$ for $l$, and $2\mu \text{F}$ for $C$ in equation (VI) to find $V\text{'}$.

  $\begin{array}{c}E\text{'}=\frac{2.0\times {10}^{-4}\text{C}}{\left(2\text{μF}\times \frac{1\text{F}}{1\times {10}^6\text{μF}}\right){10}^{-3}\text{m}}\\ =1.0\times {10}^5\text{V}\cdot {\text{m}}^{-1}\end{array}$

Therefore, the electric field between the plates of the capacitor on insertion of dielectric is $\underset{\_}{1.0\times {10}^5\text{V}\cdot {\text{m}}^{-1}}$.