A PAPER-FILLED CAPACITOR GOAL Calculate fundamental physical properties of a parallel-plate capacitor with a dielectric. PROBLEM A parallel-plate capacitor has plates 2.0 cm by 3.0 cm. The plates are separated by a 1.0-mm thickness of paper. Find (a) the capacitance of this device and (b) the

Question

A PAPER-FILLED CAPACITOR

GOAL Calculate fundamental physical properties of a parallel-plate capacitor with a dielectric.

PROBLEM A parallel-plate capacitor has plates [latex]2.0 \mathrm{~cm}[/latex] by [latex]3.0 \mathrm{~cm}[/latex]. The plates are separated by a [latex]1.0[/latex]-mm thickness of paper. Find (a) the capacitance of this device and (b) the maximum charge that can be placed on the capacitor. (c) After the fully charged capacitor is disconnected from the battery, the dielectric is subsequently removed. Find the new electric field across the capacitor. Does the capacitor discharge?

STRATEGY For part (a), obtain the dielectric constant for paper from Table [latex]16.1[/latex] and substitute, with other given quantities, into Equation 16.19.

[latex]C =\kappa \epsilon_{0} \begin{array}{l}A \ \hline d\end{array}[/latex] [16.19]

For part (b), note that Table [latex]16.1[/latex]

Table 16.1 Dielectric Constants and Dielectric Strengths
of Various Materials at Room Temperature
[latex]\begin{array}{lcc} \hline ext { Material } & \begin{array}{c} ext { Dielectric } \ ext { Constant } \kappa \end{array} & \begin{array}{c} ext { Dielectric } \ ext { Strength } (\mathbf{V} / \mathbf{m}) \end{array} \\hline ext { Air } & 1.00059 & 3 imes 10^6 \ ext { Bakelite }{ }^{\circledR} & 4.9 & 24 imes 10^6 \ ext { Fused quartz } & 3.78 & 8 imes 10^6 \ ext { Neoprene rubber } & 6.7 & 12 imes 10^6 \ ext { Nylon } & 3.4 & 14 imes 10^6 \ ext { Paper } & 3.7 & 16 imes 10^6 \ ext { Polystyrene } & 2.56 & 24 imes 10^6 \ ext { Pyrex }{ }^{\circledR} ext { glass } & 5.6 & 14 imes 10^6 \ ext { Silicone oil } & 2.5 & 15 imes 10^6 \ ext { Strontium titanate } & 233 & 8 imes 10^6 \ ext { Teflon } & 2.1 & 60 imes 10^6 \ ext { Vacuum } & 1.00000 & - \ ext { Water } & 80 & - \ \hline \end{array}[/latex]

also gives the dielectric strength of paper, which is the maximum electric field that can be applied before electrical breakdown occurs. Use Equation 16.3, [latex] \Delta V=E d[/latex], to obtain the maximum voltage and substitute into the basic capacitance equation. For part (c), remember that disconnecting the battery traps the extra charge on the plates, which must remain even after the dielectric is removed. Find the charge density on the plates and use Gauss' law to find the new electric field between the plates.

Accepted Answer

(a) Find the capacitance of this device.

Substitute into Equation 16.19:

[latex]\begin{aligned}C &=\kappa \epsilon_{0} \begin{array}{l}A \ \hline d\end{array} \&=3.7\left(\begin{array}{ll}8.85 imes 10^{-12} \frac{\mathrm{C}^{2}} {\mathrm{~N} \cdot \mathrm{m}^{2}}\end{array} ight)\left(\begin{array}{c}6.0 imes 10^{-4} \mathrm{~m}^{2} \ \hline 1.0 imes 10^{-3} \mathrm{~m}\end{array} ight) \&=2.0 imes 10^{-11} \mathrm{~F}\end{aligned}[/latex]

(b) Find the maximum charge that can be placed on the capacitor.

Calculate the maximum applied voltage, using the dielectric strength of paper, [latex]E_{\max }[/latex] :

[latex]\begin{aligned}\Delta V_{\max } &=E_{\max } d=\left(16 imes 10^{6} \mathrm{~V} / \mathrm{m} ight)\left(1.0 imes 10^{-3} \mathrm{~m} ight) \&=1.6 imes 10^{4} \mathrm{~V} \end{aligned}[/latex]

Solve the basic capacitance equation for [latex]Q_{\max }[/latex] and substitute [latex]\Delta V_{\max }[/latex] and [latex]C[/latex] :

[latex]\begin{aligned}Q_{\max } &=C \Delta V_{\max }=\left(2.0 imes 10{ }^{-11} \mathrm{~F} ight)\left(1.6 imes 10^{4} \mathrm{~V} ight) \&=0.32 \mu \mathrm{C}\end{aligned}[/latex]

(c) Suppose the fully charged capacitor is disconnected from the battery and the dielectric is subsequently removed. Find the new electric field between the plates of the capacitor.

Does the capacitor discharge?

[latex]\begin{aligned}&\sigma=\frac{Q_{m_{2 \mathrm{X}}}}{A}=\frac{3.2 imes 10^{-7} \mathrm{C}}{6.0 imes 10^{-4} \mathrm{~m}^{2}}=5.3 imes 10{ }^{-4} \mathrm{C} / \mathrm{m}^{2} \end{aligned}[/latex]

Compute the charge density on the plates:

[latex]\begin{aligned}&E=\frac{\sigma}{\epsilon_{0}}=\frac{5.3 imes 10^{-4} \mathrm{C} / \mathrm{m}^{2}}{8.85 imes 10^{-12} \mathrm{C}^{2} / \mathrm{m}^{2} \cdot \mathrm{N}}=6.0 imes 10^{7} \mathrm{~N} / \mathrm{C}\end{aligned}[/latex]

Because the electric field without the dielectric exceeds the value of the dielectric strength of air, the capacitor discharges across the gap.

REMARKS Dielectrics allow [latex]\kappa[/latex] times as much charge to be stored on a capacitor for a given voltage. They also allow an increase in the applied voltage by increasing the threshold of electrical breakdown.

Question 16.11: A PAPER-FILLED CAPACITOR GOAL Calculate fundamental physical...

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