Question

In: Physics

You need to construct a 200 pF capacitor for a science project. You plan to cut two L x L metal squares and place spacers between them. The thinnest spacers you have are 0.20 mm thick.

You need to construct a 200 pF capacitor for a science project. You plan to cut two L x L metal squares and place spacers between them. The thinnest spacers you have are 0.20 mm thick.

Part A - What is the proper value of L?
Express your answer using two significant figures.
L = _________cm

 

Solutions

Expert Solution

Concepts and reason

The concepts needed to solve this problem is the equation for the capacitance of a parallel plate capacitor. Write the expression for the capacitance of a parallel plate capacitor. Use an equation for the area of a square in the capacitance equation and solve for the square's length.

Fundamentals

The capacitance of a parallel plate capacitor can be calculated using the following formula:

\(C=\frac{\varepsilon_{0} A}{d}\)

Here, \(\varepsilon_{0}\) is the permittivity of free space and \(\mathrm{A}\) is the area of each plate of the capacitor, and \(\mathrm{d}\) is the space between the two plates. The area of a square plate of the length of each side \(L\) is, \(A=L^{2}\)

 

The capacitance of a parallel plate capacitor is,

$$ C=\frac{\varepsilon_{0} A}{d} $$

The area of a square plate of the length of each side \(L\) is,

$$ A=L^{2} $$

Substitute \(L^{2}\) for \(\mathrm{A}\) in the equation \(C=\frac{\varepsilon_{0} A}{d}\) \(C=\frac{\varepsilon_{0} L^{2}}{d}\)

Rearrange the equation for L. \(L=\sqrt{\frac{C d}{\varepsilon_{0}}}\)

The equation for the length of each side of the square plates of the capacitor is derived in terms of all given variables.

 

The final expression for length of each side of square plates of capacitor is,

$$ L=\sqrt{\frac{C d}{\varepsilon 0}} $$

Convert the unit of capacitance from pF to Farad.

$$ \begin{aligned} C &=200 \mathrm{pF}\left(\frac{10^{-12} \mathrm{~F}}{1 \mathrm{pF}}\right) \\ &=200 \times 10^{-12} \mathrm{~F} \end{aligned} $$

Convert the unit of separation between two plates from \(\mathrm{mm}\) to \(\mathrm{m}\).

$$ \begin{aligned} d &=0.20 \mathrm{~mm}\left(\frac{1 \mathrm{~m}}{1000 \mathrm{~mm}}\right) \\ &=0.20 \times 10^{-3} \mathrm{~m} \end{aligned} $$

Substitute \(200 \times 10^{-12} \mathrm{~F}\) for \(\mathrm{C}, 8.854 \times 10^{-12} \mathrm{C}^{2} / \mathrm{N} \cdot \mathrm{m}^{2}\) for \(\varepsilon_{0},\) and \(0.20 \times 10^{-3} \mathrm{~m}\) for \(\mathrm{d}\) in the equation

\(L=\sqrt{\frac{C d}{\varepsilon_{0}}}\)

$$ \begin{array}{c} L=\sqrt{\frac{\left(200 \times 10^{-12} \mathrm{~F}\right)\left(0.20 \times 10^{-3} \mathrm{~m}\right)}{8.854 \times 10^{-12} \mathrm{C}^{2} / \mathrm{N} \cdot \mathrm{m}^{2}}} \\ =0.0672 \mathrm{~m}\left(\frac{100 \mathrm{~cm}}{1 \mathrm{~m}}\right) \\ =6.7 \mathrm{~cm} \end{array} $$

The length of each side of the capacitor (in two significant digits) is \(6.7 \mathrm{~cm} .\)

The value of permittivity of free space is equal to \(8.854 \times 10^{-12} \mathrm{C}^{2} / \mathrm{N} \cdot \mathrm{m}^{2} .\) The required value of length of each side of the square plates of capacitor is calculated by substitute values of all the variables in same system of units and the final answer is given in two significant digits in required units.

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