Chapter 23: Problem 18
The electric field just above the surface of the charged conducting drum of a photocopying machine has a magnitude \(E\) of \(2.3 \times 10^{5} \mathrm{~N} / \mathrm{C} .\) What is the surface charge density on the drum?
Short Answer
Expert verified
The surface charge density is approximately \(2.036 \times 10^{-6} \mathrm{~C/m^2}\).
Step by step solution
01
Understanding the Concept
To find the surface charge density, we will apply Gauss’s Law. For a charged conductor, the electric field just outside the surface of the conductor is related to the surface charge density \(\sigma\) by the equation \(E = \frac{\sigma}{\varepsilon_0}\), where \(\varepsilon_0\) is the permittivity of free space and equals approximately \(8.85 \times 10^{-12} \mathrm{~C^2/N \cdot m^2}\).
02
Substitute Known Values
We know the electric field \(E = 2.3 \times 10^5 \mathrm{~N/C}\). The equation \(E = \frac{\sigma}{\varepsilon_0}\) can be rearranged to find the surface charge density: \(\sigma = E \times \varepsilon_0\).
03
Calculate Surface Charge Density
Substitute \(E = 2.3 \times 10^5 \mathrm{~N/C}\) and \(\varepsilon_0 = 8.85 \times 10^{-12} \mathrm{~C^2/N \cdot m^2}\) into the equation:\[\sigma = (2.3 \times 10^{5} \mathrm{~N/C}) \times (8.85 \times 10^{-12} \mathrm{~C^2/N \cdot m^2})\]This gives \(\sigma \approx 2.036 \times 10^{-6} \mathrm{~C/m^2}\).
04
Final Answer
The surface charge density on the drum is approximately \(2.036 \times 10^{-6} \mathrm{~C/m^2}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Electric Field
An electric field describes the force a charge experiences per unit charge around any charged object. It's a way to visualize how charges influence each other from a distance. In simpler terms, imagine the electric field like an invisible bubble that surrounds charged objects. If you put a positive test charge in this field, it will experience a force. By convention, the direction of the electric field is the direction a positive test charge would be pushed or pulled.
The strength of the electric field, denoted as \(E\), is measured in newtons per coulomb (\(\mathrm{N/C}\)). In the exercise, the electric field just above the surface of a charged drum is given as \(2.3 \times 10^{5} \mathrm{~N/C}\). This indicates a very strong field, capable of exerting significant force on any nearby charges. The electric field is not just a theoretical idea; it is very practical, influencing everything from simple household electronics to large-scale industrial equipment like photocopiers.
The strength of the electric field, denoted as \(E\), is measured in newtons per coulomb (\(\mathrm{N/C}\)). In the exercise, the electric field just above the surface of a charged drum is given as \(2.3 \times 10^{5} \mathrm{~N/C}\). This indicates a very strong field, capable of exerting significant force on any nearby charges. The electric field is not just a theoretical idea; it is very practical, influencing everything from simple household electronics to large-scale industrial equipment like photocopiers.
Gauss's Law
Gauss's Law is a vital principle in electromagnetism that links the electric field to the charge distribution causing it. In basic terms, it states that the net electric flux passing through a closed surface is equal to the charge enclosed divided by the permittivity of free space (\(\varepsilon_0\)). It is usually expressed mathematically as:
For the charged drum in a copier, Gauss's Law tells us that the electric field just outside the surface of a conductor is directly proportional to the surface charge density \(\sigma\) divided by \(\varepsilon_0\). This relationship is vital for calculations involving metals and conductive materials. In the exercise, recognizing this allowed us to derive \( \sigma = E \times \varepsilon_0 \), which gives direct insight into the surface charge density from a known electric field.
- \( \Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\varepsilon_0} \)
For the charged drum in a copier, Gauss's Law tells us that the electric field just outside the surface of a conductor is directly proportional to the surface charge density \(\sigma\) divided by \(\varepsilon_0\). This relationship is vital for calculations involving metals and conductive materials. In the exercise, recognizing this allowed us to derive \( \sigma = E \times \varepsilon_0 \), which gives direct insight into the surface charge density from a known electric field.
Permittivity of Free Space
The permittivity of free space, symbolized as \(\varepsilon_0\), is a fundamental constant in physics that plays a crucial role in electromagnetism. It describes how electric fields interact with and are affected by the vacuum of space. Essentially, \(\varepsilon_0\) measures the resistance encountered when forming an electric field in a vacuum.
Its value is approximately \(8.85 \times 10^{-12} \mathrm{~C^2/N \cdot m^2}\), and it appears in many fundamental equations like Gauss's Law and Coulomb's Law. In the context of the exercise, \(\varepsilon_0\) is part of the formula \(E = \frac{\sigma}{\varepsilon_0}\), which allows us to determine the surface charge density from a known electric field. It shows the proportional relationship between the electric field strength and surface charge density. Understanding \(\varepsilon_0\) helps us grasp how electrical interactions behave in the simplest environments, forming a baseline for more complex scenarios and materials.
Its value is approximately \(8.85 \times 10^{-12} \mathrm{~C^2/N \cdot m^2}\), and it appears in many fundamental equations like Gauss's Law and Coulomb's Law. In the context of the exercise, \(\varepsilon_0\) is part of the formula \(E = \frac{\sigma}{\varepsilon_0}\), which allows us to determine the surface charge density from a known electric field. It shows the proportional relationship between the electric field strength and surface charge density. Understanding \(\varepsilon_0\) helps us grasp how electrical interactions behave in the simplest environments, forming a baseline for more complex scenarios and materials.