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Expert-verified Found in: Page 699 ### College Physics (Urone)

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 $$0.452\;nJ$$.

See the step by step solution

## Step 1: Concepts and Principles

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 $$\Delta V$$: $${U_E} = \frac{1}{2}C{(\Delta V)^2}...(3)$$.

## Step 2: Given Data and required data

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: $$Q = 2.00\;nC$$.

## Step 3: Finding Capacitance

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 $$4.43{\rm{ }}nF$$.

## Step 4: Finding potential difference

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 $$0.452\;V$$.

## Step 5: Finding energy stored

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 $$0.452\;nJ$$. ### Want to see more solutions like these? 