Q19.5-56PE
Question
(a) A certain parallel plate capacitor has plates of area \(4.00\;{m^2}\), separated by \(0.01\;mm\) of nylon, and stores \(0.170\;C\) of charge. What is the applied voltage? (b) What is unreasonable about this result? (c) Which assumptions are responsible or inconsistent?
Step-by-Step Solution
VerifiedThe required solution is
a) The voltage applied is \(14{\rm{ }}kV\).
(b) The voltage across the capacitor is too high.
(c) The charge on the capacitor is too high for a capacitor with this geometry.
A capacitor is a device used to store electrical energy and works in an electric field. It is a passive electrical component with two terminals. Capacitance is the term used to describe the effect of a capacitor.
Any pair of conductors separated by an insulating substance is referred to as a capacitor. When the capacitor is charged, the two conductors have charges of equal magnitude \(Q\) and opposite sign, and the positively charged conductor's potential \(\Delta V\) with respect to the negatively charged conductor is proportional to \(Q\) The ratio of \(Q\) to \(\Delta V\) determines the capacitance \(C\)
\(C = \frac{Q}{{\Delta V}}\)
The farad is the SI unit of capacitance (\(F\)): \(F = 1C/V\)
The parallel plate capacitor's capacitance is \(C = \kappa \frac{{{\varepsilon _0}A}}{d}\)
Where \(A\) is the area of each plate, \({\rm{d}}\)is the distance between them, and \({\varepsilon _0}\)is the vacuum permittivity
\(\begin{align}{\underline{\phantom{xx}}}{\varepsilon _0} &= 1/(4\pi k)\\ &= 8.854 \times {10^{ - 12}}{\rm{ }}{C^2}/\left( {N \times {m^2}} \right)\end{align}\)
If the plates are separated by vacuum, \(\kappa = 1\); otherwise, the dielectric constant of the dielectric is \(\kappa > 1\).
In a homogeneous electric field of magnitude \(E\), the potential difference between two locations separated by a distance \(d\) is
The plates of the capacitor have the following area: \(A = 4.00\;{m^2}\)
The capacitor plates are separated by the following distance:
\(\begin{align}{\underline{\phantom{xx}}}d & = (0.0100\;mm)\left( {\frac{{1\;m}}{{1000\;mm}}} \right)\\ & = 0.0100 \times {10^{ - 3}}\;m\end{align}\)
Neoprene has a dielectric constant of: \(\kappa = 3.4\)
Across the capacitor, the potential difference is: \(\Delta V = Ed\)
The capacitor holds the following charge: \(Q = 0.170{\rm{ }}C\)
Equation is used to calculate the parallel plate capacitor's capacitance.
\(C = \kappa \frac{{{\varepsilon _0}A}}{d}\)
Substitute the values of \(\kappa \),\({\varepsilon _0}\),\(A\) and \(d\) -
\(\begin{align}{\underline{\phantom{xx}}}C & = (3.4)\left[ {8.854 \times {{10}^{ - 12}}{\rm{ }}{C^2}/\left( {N \times {m^2}} \right)} \right]\frac{{4.00\;{m^2}}}{{0.0100 \times {{10}^{ - 3}}\;m}}\\ & = 1.2 \times {10^{ - 5}}\;F\end{align}\)
Therefore, the parallel plate capacitor's capacitance is \(1.2 \times {10^{ - 5}}\;F\).
a) Equation is used to calculate the potential difference across the capacitor.
\(\Delta V = \frac{Q}{C}\)
Substitute the values of \(Q\) and \(C\):
\(\begin{align}{\underline{\phantom{xx}}}\,\Delta V & = \frac{{0.170{\rm{ }}C}}{{1.2 \times {{10}^{ - 5}}\;F}}\\ & = 14 \times {10^3}\;V\\ & = \left( {14 \times {{10}^3}\;V} \right)\left( {\frac{{1{\rm{ }}kV}}{{{{10}^3}\;V}}} \right)\\ & = 14{\rm{ }}kV\end{align}\)
Therefore, the voltage applied is \(14{\rm{ }}kV\).
b) Because nylon has a dielectric strength of \(14 \times {10^6}\;V/m\), the maximum applied voltage for a \(0.0100\;mm\) separation filled with nylon is
\(\begin{align}{\underline{\phantom{xx}}}V & = Ed\\ & = \left( {14 \times {{10}^6}\;V/m} \right)\left( {0.0100 \times {{10}^{ - 3}}\;m} \right)\\ & = 140\;V\end{align}\)
As a result, the voltage in portion (a) is \(100\) times higher than the maximum voltage that can be supplied to a capacitor of this type.
c) For a capacitor with this geometry, the charge on the capacitor is too high.