Chapter 17, Problem 76P

(a)

To determine

The maximum stored charge on the capacitor.

Answer to Problem 76P

The maximum stored charge on the capacitor is $200\text{nC}$.

Explanation of Solution

The diameter of the circular plate is $10.0\text{cm}$ and the separation between the plates is $2.00\text{mm}$.

Write the expression for charge stored on each plate of the capacitor

$Q_{\text{m}}=C\Delta V_{\text{m}}$                                                                        (I)

Here, $Q_{\text{m}}$ is the magnitude of maximum charge stored on each plate of the capacitor, $C$ is the capacitance of the capacitor and $\Delta V_{\text{m}}$ is the potential difference corresponding to dielectric breakdown.

Write the expression for capacitance of parallel plate capacitor

$C=\frac{A}{4\pi kd}$                                                                          (II)

Here, $C$ is the capacitance of the capacitor, $k$ is the Coulomb’s constant, $A$ is the area of the plates and $d$ is the distance between the plates.

Write the expression for area of the circular plate

$A=\frac{\pi D^2}{4}$                                                                           (III)

Here, $D$ is the diameter of the plate.

Substitute (III) in (II)

$C=\frac{\pi D^2}{16\pi kd}$

Cancel the common terms

$C=\frac{D^2}{16kd}$                                                                            (IV)

Write the expression for potential difference

$\Delta V_{\text{m}}=E_{\text{m}}d$                                                                          (V)

Here, $E_{\text{m}}$ is the dielectric strength.

Substitute (IV) and (V) in (I)

$Q_{\text{m}}=\frac{D^2}{16kd}\left(E_{\text{m}}d\right)$

Rearrange

$Q_{\text{m}}=\frac{D^2}{16k}\left(E_{\text{m}}\right)$                                                                  (VI)

The dielectric strength of dry air is $3\times {10}^6{\text{Vm}}^{-1}$.

Substitute $8.988\times {10}^9{\text{NC}}^{-2}{\text{m}}^2$ for $k$, $10.0\text{cm}$ for $D$  and $3\times {10}^6{\text{Vm}}^{-1}$ for $E_{\text{m}}$ in (VI) to find $Q_{\text{m}}$

$\begin{array}{c}{Q}_{\text{m}}=\frac{{\left(10.0\text{cm}\right)}^2\left(3\times {10}^6{\text{Vm}}^{-1}\right)}{16\left(8.988\times {10}^9{\text{NC}}^{-2}{\text{m}}^2\right)}\\ =\frac{{\left(10.0\times {10}^{-2}\text{m}\right)}^2\left(3\times {10}^6{\text{Vm}}^{-1}\right)}{16\left(8.988\times {10}^9{\text{NC}}^{-2}{\text{m}}^2\right)}\cdot \frac{\text{1}{\text{NmV}}^{-1}}{\text{1}\text{C}}\\ =200\times {10}^{-9}\text{C}\\ =200\text{nC}\end{array}$

Thus, the maximum stored charge on the capacitor is $200\text{nC}$.

(b)

To determine

The maximum stored charge on the capacitor with neoprene between the plates.

Answer to Problem 76P

The maximum stored charge on the capacitor with neoprene between the plates is $5.6\text{μC}$.

Explanation of Solution

The diameter of the circular plate is $10.0\text{cm}$ and the separation between the plates is $2.00\text{mm}$.

Write the expression for charge stored on each plate of the capacitor with a dielectric

${Q^{\prime }}_{\text{m}}=C^{\prime }\Delta V_{\text{m}}$                                                                                     (VII)

Here, ${Q^{\prime }}_{\text{m}}$ is the magnitude of maximum charge stored on each plate of the capacitor with a dielectric and $C^{\prime }$ is the capacitance of the capacitor with a dielectric.

Write the expression for capacitance of parallel plate capacitor with a dielectric

$C^{\prime }=\kappa C$                                                                                         (VIII)

Here, $\kappa$ is the dielectric constant of neoprene.

Substitute (IV) in (VIII)

$C^{\prime }=\kappa \frac{D^2}{16kd}$                                                                                   (IX)

Substitute (IX) and (V) in (VII)

${Q^{\prime }}_{\text{m}}=\kappa \frac{D^2}{16kd}\left(E_{\text{m}}d\right)$                                                                        (X)

The dielectric strength of neoprene is $12\times {10}^6{\text{Vm}}^{-1}$ and its dielectric constant is $6.7$.

Substitute $8.988\times {10}^9{\text{NC}}^{-2}{\text{m}}^2$ for $k$, $10.0\text{cm}$ for $D$  and $12\times {10}^6{\text{Vm}}^{-1}$ for $E_{\text{m}}$ in (X) to find ${Q^{\prime }}_{\text{m}}$

$\begin{array}{c}{Q^{\prime }}_{\text{m}}=\frac{6.7{\left(10.0\text{cm}\right)}^2\left(12\times {10}^6{\text{Vm}}^{-1}\right)}{16\left(8.988\times {10}^9{\text{NC}}^{-2}{\text{m}}^2\right)}\\ =\frac{6.7{\left(10.0\times {10}^{-2}\text{m}\right)}^2\left(12\times {10}^6{\text{Vm}}^{-1}\right)}{16\left(8.988\times {10}^9{\text{NC}}^{-2}{\text{m}}^2\right)}\cdot \frac{\text{1}{\text{NmV}}^{-1}}{\text{1}\text{C}}\\ =5.6\times {10}^{-6}\text{C}\\ =5.6\text{μC}\end{array}$

Thus, the maximum stored charge on the capacitor with a dielectric is $5.6\text{μC}$.