Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration


College Physics (Urone)
Found in: Page 699

Answers without the blur.

Just sign up for free and you're in.


Short Answer

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 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


  • 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


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 


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  


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 \(0.452\;nJ\).

Recommended explanations on Physics Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.