Question 16.11: A PAPER-FILLED CAPACITOR GOAL Calculate fundamental physical...
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 \mathrm{~cm} by 3.0 \mathrm{~cm}. The plates are separated by a 1.0-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 16.1 and substitute, with other given quantities, into Equation 16.19.
C =\kappa \epsilon_{0} \begin{array}{l}A \\ \hline d\end{array} [16.19]
For part (b), note that Table 16.1
Table 16.1 Dielectric Constants and Dielectric Strengths
of Various Materials at Room Temperature
\begin{array}{lcc} \hline \text { Material } & \begin{array}{c} \text { Dielectric } \\ \text { Constant } \kappa \end{array} & \begin{array}{c} \text { Dielectric } \\ \text { Strength } (\mathbf{V} / \mathbf{m}) \end{array} \\\hline \text { Air } & 1.00059 & 3 \times 10^6 \\ \text { Bakelite }{ }^{\circledR} & 4.9 & 24 \times 10^6 \\ \text { Fused quartz } & 3.78 & 8 \times 10^6 \\ \text { Neoprene rubber } & 6.7 & 12 \times 10^6 \\ \text { Nylon } & 3.4 & 14 \times 10^6 \\ \text { Paper } & 3.7 & 16 \times 10^6 \\ \text { Polystyrene } & 2.56 & 24 \times 10^6 \\ \text { Pyrex }{ }^{\circledR} \text { glass } & 5.6 & 14 \times 10^6 \\ \text { Silicone oil } & 2.5 & 15 \times 10^6 \\ \text { Strontium titanate } & 233 & 8 \times 10^6 \\ \text { Teflon } & 2.1 & 60 \times 10^6 \\ \text { Vacuum } & 1.00000 & – \\ \text { Water } & 80 & – \\ \hline \end{array}
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, \Delta V=E d, 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.
(a) Find the capacitance of this device.
Substitute into Equation 16.19:
\begin{aligned}C &=\kappa \epsilon_{0} \begin{array}{l}A \\ \hline d\end{array} \\&=3.7\left(\begin{array}{ll}8.85 \times 10^{-12} \frac{\mathrm{C}^{2}} {\mathrm{~N} \cdot \mathrm{m}^{2}}\end{array}\right)\left(\begin{array}{c}6.0 \times 10^{-4} \mathrm{~m}^{2} \\ \hline 1.0 \times 10^{-3} \mathrm{~m}\end{array}\right) \\&=2.0 \times 10^{-11} \mathrm{~F}\end{aligned}
(b) Find the maximum charge that can be placed on the capacitor.
Calculate the maximum applied voltage, using the dielectric strength of paper, E_{\max } :
\begin{aligned}\Delta V_{\max } &=E_{\max } d=\left(16 \times 10^{6} \mathrm{~V} / \mathrm{m}\right)\left(1.0 \times 10^{-3} \mathrm{~m}\right) \\&=1.6 \times 10^{4} \mathrm{~V} \end{aligned}
Solve the basic capacitance equation for Q_{\max } and substitute \Delta V_{\max } and C :
\begin{aligned}Q_{\max } &=C \Delta V_{\max }=\left(2.0 \times 10{ }^{-11} \mathrm{~F}\right)\left(1.6 \times 10^{4} \mathrm{~V}\right) \\&=0.32 \mu \mathrm{C}\end{aligned}
(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?
\begin{aligned}&\sigma=\frac{Q_{m_{2 \mathrm{X}}}}{A}=\frac{3.2 \times 10^{-7} \mathrm{C}}{6.0 \times 10^{-4} \mathrm{~m}^{2}}=5.3 \times 10{ }^{-4} \mathrm{C} / \mathrm{m}^{2} \end{aligned}
Compute the charge density on the plates:
\begin{aligned}&E=\frac{\sigma}{\epsilon_{0}}=\frac{5.3 \times 10^{-4} \mathrm{C} / \mathrm{m}^{2}}{8.85 \times 10^{-12} \mathrm{C}^{2} / \mathrm{m}^{2} \cdot \mathrm{N}}=6.0 \times 10^{7} \mathrm{~N} / \mathrm{C}\end{aligned}
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 \kappa 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.