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Q67PE

Expert-verifiedFound in: Page 699

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**A nervous physicist worries that the two metal shelves of his wood frame bookcase might obtain a high voltage if charged by static electricity, perhaps produced by friction. **

**(a) What is the capacitance of the empty shelves if they have area \(1.00 \times {10^2}\;{m^2}\) and are \(0.200\;m\) apart? **

**(b) What is the voltage between them if opposite charges of magnitude \(2.00nc\) are placed on them? **

**(c) To show that this voltage poses a small hazard, calculate the energy stored.**

The capacitance of the empty shelves if they have area \(1.00 \times {10^2}\;{m^2}\) and are \(0.200\;m\) apart is \(4.43{\rm{ }}nF\).

b) The voltage between them if opposite charges of magnitude \(2.00{\rm{ }}nc\) are placed is \(0.452\;V\).

c) The energy stored is \(0.452\;nJ\).

The parallel plate capacitor's capacitance is \(C = \kappa \frac{{{\varepsilon _0}A}}{d}...(1)\)

Where \(A\) is the area of each plate, \({\rm{d}}\)is the distance between them, and \({\varepsilon _0}\)is the vacuum permittivity

\(\begin{aligned}{}{\varepsilon _0} &= 1/(4\pi k)\\ = 8.854 \times {10^{ - 12}}{\rm{ }}{C^2}/\left( {N \times {m^2}} \right)\end{aligned}\)

If the plates are separated by vacuum, \(\kappa = 1\); otherwise, the dielectric constant of the dielectric is \(\kappa > 1\).

**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 **\(C\).

\(C = \frac{Q}{{\Delta V}}....(2)\)

The farad is the SI unit of capacitance (\(F\)): \(F = 1C/V\).

** **

**The following is the amount of energy held in a capacitor with capacitance **\(C\)** that is charged to a potential difference **\(\Delta V\)**: **\({U_E} = \frac{1}{2}C{(\Delta V)^2}...(3)\)**.**

Given

- The area of each plate is: \(A = 1.00 \times {10^2}\;\;{m^2}\).
- The separation between the plates is: \(d = 0.200\;\;m\).
- The charge stored on the metal shelves is: \(Q = 2.00\;nC\).

a)

The capacitance of the metal shelve is found from Equation \((1)\):

\(C = \kappa \frac{{{\varepsilon _0}A}}{d}\)

Where \(\kappa = 1\), because the insulating material is air.

Substitute the rest of numerical values from given data:

\(\begin{aligned}{}C &= (1)\left( {8.854 \times {{10}^{ - 12}}{\rm{ }}{C^2}/\left( {N \times {m^2}} \right)} \right)\frac{{1.00 \times {{10}^2}\;{m^2}}}{{0.200\;m}}\\ &= 4.43 \times {10^{ - 9}}\;F\\ &= \left( {4.43 \times {{10}^{ - 9}}\;F} \right)\left( {\frac{{{{10}^9}{\rm{ }}nF}}{{1\;F}}} \right)\\ &= 4.43{\rm{ }}nF\end{aligned}\)

Therefore, the capacitance value is obtained as \(4.43{\rm{ }}nF\).

b)

The potential difference between the two shelves is found by solving Equation \((2)\) for \(\Delta V\):

\(\Delta V = \frac{Q}{C}\)

Entering the values for \(Q\) and \(C\), we obtain:

\(\begin{aligned}{}\Delta V &= \frac{{2.00{\rm{ }}nF}}{{4.43{\rm{ }}nF}}\\ &= 0.452\;V\end{aligned}\)

Hence, the voltage value is obtained as \(0.452\;V\).

c)

The energy stored in the capacitor is found from Equation \((3)\):

\({U_E} = \frac{1}{2}C{(\Delta V)^2}\)

Substitute numerical values:

\(\begin{aligned}{}{U_E} &= \frac{1}{2}\left( {4.43 \times {{10}^{ - 9}}\;F} \right){(0.452\;V)^2}\\ &= 0.542 \times {10^{ - 9}}\;J\\ &= \left( {0.542 \times {{10}^{ - 9}}\;J} \right)\left( {\frac{{{{10}^9}{\rm{ }}nJ}}{{1\;J}}} \right)\\ &= 0.452\;nJ\end{aligned}\)

Hence, the energy value is obtained as \(0.452\;nJ\).

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