Chapter 21, Problem 99A

(a)

To determine

The quantity of charge stored in the capacitor.

Answer to Problem 99A

  $q=5.6\times {\text{10}}^{-6}\text{ C}$

Explanation of Solution

Given:

Capacitance of a capacitor is $C=0.047\mu \text{F = 0}\text{.047}\times {\text{10}}^{-6}\text{ F}$

The separation between the capacitor plates is $d\text{ = 0}\text{.25 cm = 0}\text{.25}\times {\text{10}}^{-2}\text{ m}$

Potential difference between the plates is $\Delta \text{V = 120 V}$

Formula used:

The charge in the capacitor can be calculated using the equation,

  $q=C\Delta V$ $\left(1\right)$

Where, $C$ is the capacitance of a capacitor,

  $\Delta V$ is the potential difference between the capacitor plates

Calculation:

Substituting the numerical values in equation $\left(1\right)$ ,

  $q=\left(\text{0}\text{.047}\times {\text{10}}^{-6}\text{ F}\right)\left(\text{120 V}\right)$

  $=5.6\times {\text{10}}^{-6}\text{ C}$

Conclusion:

The charge stored in the capacitor is $5.6\times {\text{10}}^{-6}\text{ C}$ .

(b)

To determine

The strength of the electric field between the plates of the capacitor.

Answer to Problem 99A

  $E=4.8\times {10}^4\text{ V/m}$

Explanation of Solution

Given:

Capacitance of a capacitor is $C=0.047\mu \text{F = 0}\text{.047}\times {\text{10}}^{-6}\text{ F}$

The separation between the capacitor plates is $d\text{ = 0}\text{.25 cm = 0}\text{.25}\times {\text{10}}^{-2}\text{ m}$

Potential difference between the plates is $\Delta \text{V = 120 V}$

Formula used:

Potential difference between the capacitor plates is,

  $\Delta V=Ed$

  $\Rightarrow E=\frac{\Delta V}{d}$ $\left(2\right)$

Where, $E$ is the electric field strength

  $d$ separation between the capacitor plates

Calculation:

Substituting the numerical values in equation $\left(2\right)$ ,

  $E=\frac{\text{120 V}}{\text{0}\text{.25}\times {\text{10}}^{-2}\text{ m}}=4.8\times {10}^4\text{ V/m}$

Conclusion:

The strength of the electric field between the plates of the capacitor is $4.8\times {10}^4\text{ V/m}$ .

(c)

To determine

Force exerted on an electron placed between the plates of a capacitor.

Answer to Problem 99A

Force exerted on an electron placed between the plates of a capacitor is $7.68\times {\text{10}}^{-15}\text{ N}$

Explanation of Solution

Given:

Capacitance of a capacitor is $C=0.047\mu \text{F = 0}\text{.047}\times {\text{10}}^{-6}\text{ F}$

The separation between the capacitor plates is $d\text{ = 0}\text{.25 cm = 0}\text{.25}\times {\text{10}}^{-2}\text{ m}$

Potential difference between the plates is $\Delta \text{V = 120 V}$

The position of the electron is shown in the figure below.

Formula used:

The electric field strength $\left(E\right)$ is given by,

  $E=\frac{F}{q}$

  $\Rightarrow F=Eq$ $\left(3\right)$

Where, $F$ is the electric force

$q$ is the charge of an electron, $q=1.6\times {\text{10}}^{-19}\text{ C}$

  $E=4.8\times {10}^4\text{ V/m}$ from part (b)

Calculation:

Substituting the numerical values in equation $\left(3\right)$ ,

  $F=\left(4.8\times {10}^4\text{ V/m}\right)\left(1.6\times {\text{10}}^{-19}\text{ C}\right)$

  $=7.68\times {\text{10}}^{-15}\text{ N}$

Conclusion:

Force exerted on an electron placed between the plates of a capacitor is $7.68\times {\text{10}}^{-15}\text{ N}$ .