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Chapter 17: Problem 113

It has only been fairly recently that 1.0 -F capacitors have been readily available. A typical 1.0 -F capacitor can withstand up to \(5.00 \mathrm{V}\). To get an idea why it isn't easy to make a 1.0 -F capacitor, imagine making a \(1.0-\mathrm{F}\) parallel plate capacitor using titanium dioxide \((\kappa=90.0\) breakdown strength \(4.00 \mathrm{kV} / \mathrm{mm}\) ) as the dielectric. (a) Find the minimum thickness of the titanium dioxide such that the capacitor can withstand \(5.00 \mathrm{V}\). (b) Find the area of the plates so that the capacitance is \(1.0 \mathrm{F}\)

Short Answer

Expert verified
Answer: The minimum thickness of the titanium dioxide dielectric is 0.00125 mm, and the area of the capacitor plates is approximately 260.73 m².

Step by step solution

01

Find the minimum thickness of titanium dioxide

To find the minimum thickness of the dielectric, we need to use the breakdown strength property. The breakdown strength (B) is given in \(4.00 \mathrm{kV}/\mathrm{mm}\), which is the maximum electric field (E) it can withstand before breaking down. We also know the maximum voltage (V) is \(5.00 \mathrm{V}\). We can use the formula for electric field in a parallel plate capacitor: $$ E = \frac{V}{d} $$ where d is the distance between the plates or the thickness of the dielectric. So, we can rearrange the formula for d: $$ d = \frac{V}{E} $$ Now we can substitute the given values and convert units to find the minimum thickness.
02

Calculate the minimum thickness

We know that the capacitor must withstand a \(5.00 \mathrm{V}\) voltage and the breakdown strength is \(4.00 \mathrm{kV}/\mathrm{mm}\). But before calculating, let's first convert the breakdown strength to \(\mathrm{V}/\mathrm{mm}\): $$ B = 4.00 \mathrm{kV}/\mathrm{mm} = 4000 \mathrm{V}/\mathrm{mm} $$ Now we can calculate the minimum thickness: $$ d = \frac{5.00 \mathrm{V}}{4000 \mathrm{V}/\mathrm{mm}} = 0.00125 \mathrm{mm} $$ So the minimum thickness of the titanium dioxide must be \(0.00125 \mathrm{mm}\).
03

Calculate the area of the plates

Now, we need to find the area of the plates with a capacitance of \(1.0 \mathrm{F}\). We will use the formula for parallel plate capacitors: $$ C = \kappa \epsilon_0 \frac{A}{d} $$ where C is the capacitance, \(\kappa\) is the dielectric constant, \(\epsilon_0\) is the vacuum permittivity, A is the area of the plates, and d is the thickness of the dielectric. Rearranging for A, we get: $$ A = \frac{C \cdot d}{\kappa \epsilon_0} $$ We are given the values of C, d, and \(\kappa\), and the vacuum permittivity \(\epsilon_0 = 8.85 \times 10^{-12} \mathrm{F}/\mathrm{m}\).
04

Calculate the plate area

Now we can substitute the given values and calculate the area of the plates: $$ A = \frac{(1.0 \mathrm{F})(0.00125 \times 10^{-3} \mathrm{m})}{(90.0)(8.85 \times 10^{-12} \mathrm{F}/\mathrm{m})} = 260.73 \mathrm{m^2} $$ So the area of the plates of the capacitor must be approximately \(260.73 \mathrm{m^2}\) to achieve a capacitance of \(1.0 \mathrm{F}\) with a \(0.00125 \mathrm{mm}\) thick titanium dioxide dielectric.

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