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Chapter 16: Problem 57

An infinite non-conducting sheet has a surface charge density \(\sigma=0.1 \mu \mathrm{C} / \mathrm{m}^{2}\) on one side. How far apart are equipotential surfaces whose potentials differ by 50 volt? (a) \(8.8 \mathrm{~mm}\) (b) \(8.8 \mathrm{~cm}\) (c) \(8.8 \mu \mathrm{m}\) (d) \(8.8 \mathrm{pm}\)

Short Answer

Expert verified
The correct answer is (a) 8.8 mm.

Step by step solution

01

Identify the Electric Field from the Sheet

For an infinite non-conducting sheet with surface charge density \(\sigma\), the electric field \(E\) is given by:\[ E = \frac{\sigma}{2\varepsilon_0} \]where \( \varepsilon_0 \) is the permittivity of free space and its value is approximately \(8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2\). Given \(\sigma = 0.1 \times 10^{-6} \, \text{C/m}^2\), we can calculate \(E\).
02

Calculate the Electric Field

Substitute \(\sigma\) into the formula:\[ E = \frac{0.1 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} \]Calculate \(E\):\[ E = \frac{0.1 \times 10^{-6}}{17.7 \times 10^{-12}} = 5.65 \times 10^{3} \, \text{N/C} \]
03

Determine the Relationship Between Potential Difference and Distance

The potential difference \(\Delta V\) between two equipotential surfaces separated by distance \(d\) in a uniform electric field \(E\) is given by:\[ \Delta V = E \cdot d \]
04

Solve for Distance \(d\)

Given \(\Delta V = 50 \, \text{V}\), solve for \(d\):\[ d = \frac{\Delta V}{E} = \frac{50}{5.65 \times 10^{3}} \approx 8.85 \times 10^{-3} \, \text{m} \]
05

Convert Distance to Appropriate Units

Convert \(8.85 \times 10^{-3} \, \text{m}\) to millimeters:\[ 8.85 \times 10^{-3} \, \text{m} = 8.85 \, \text{mm} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equipotential Surfaces
Equipotential surfaces are fascinating features in the world of electric fields. Imagine them as invisible contours surrounding a charged object, where each surface represents a constant electric potential. They are akin to contour lines on a topographic map, each line signifying a constant elevation.

Some key features of equipotential surfaces include:
  • They are always perpendicular to electric field lines. This is because no work is done while moving along an equipotential surface, meaning the electric field most efficiently moves in one direction, at right angles to these surfaces.
  • The potential difference between any two equipotential surfaces is always the same, making them helpful tools in understanding electric fields.
  • In uniform electric fields, equipotential surfaces are parallel and evenly spaced. In non-uniform fields, they get denser near charges.
This means, in our problem, the spacing between any two equipotential surfaces with a voltage difference of 50 V remains consistent, allowing us to easily calculate the distance in a straightforward manner.
Surface Charge Density
Surface charge density, represented by the Greek letter \(\sigma\), is a measure of how much charge is distributed over a specific area of a surface. It is a crucial factor when studying electric fields generated by different objects.

Here鈥檚 how surface charge density plays a role in determining electric fields:
  • In our infinite non-conducting sheet, the charge density is given as \(0.1 \mu\text{C/m}^{2}\). This tells us exactly how much charge per unit area the sheet holds.
  • This value directly affects the electric field emanating from the sheet, with the formula \( E = \frac{\sigma}{2\varepsilon_0} \) indicating that a larger charge density creates a stronger electric field.
  • In practical terms, understanding surface charge density helps us predict how the field will interact with other charged objects or particles nearby.
Thus, grasping surface charge density is essential for solving problems involving electric forces and fields.
Permittivity of Free Space
When learning about electric fields, the term "permittivity of free space" often comes up. Scientifically designated as \(\varepsilon_0\), it is a constant that quantifies how much electric field is "permitted" in a vacuum. It serves an essential role in equations that describe electric forces and fields.

Key points to note about permittivity of free space include:
  • Its value is approximately \8.85 \times 10^{-12} \text{C}^{2}/\text{N} \cdot \text{m}^{2}\. This tiny number significantly impacts the determination of electric fields around charged objects.
  • In the formula that computes the electric field from the surface charge density, \(\varepsilon_0\) acts as a scaling factor. It affects how strong the field is for a given amount of surface charge density.
  • Beyond just calculations, \(\varepsilon_0\) is also foundational in Maxwell's equations, which are pivotal for describing classical electromagnetism.
Therefore, understanding \(\varepsilon_0\) gives deeper insight into how electric fields behave in different mediums, making it indispensable in physics.

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Most popular questions from this chapter

Gauss' law in electrostatics is true, when the charges enclosed in the Gaussian surface are (a) stationary only (b) moving only (c) moving or stationary (d) none of these

Choose the wrong statement (a) The electric field calculated by Gauss' law is the field due to the charges inside the Gaussian surface. (b) The electric field calculated by Gauss' law is the resultant field due to all the charges, inside and outside the closed surface. (c) The Gauss' law is equivalent to Coulomb's law. (d) The Gauss' law can also be applied to calculate gravitational field but with some modifications.

What is not true for equipotential surface for uniform electric field? (a) Equipotential surface is flat (b) Equipotential surface is spherical (c) Electric lines are perpendicular to equipotential surface (d) Work done is zero

Consider a neutral conducting sphere. A positive point charge is placed outside the sphere. The net charge on the sphere is then (a) negative and distributed uniformly over the surface of the sphere (b) negative and appears only at the point on the sphere closest to the point charge (c) negative and distributed non-uniformly over the entire surface of the sphere (d) zero

In a region of space the electric field is given by \(\vec{E}=\) \(8 \hat{i}+4 \hat{j}+3 \hat{k}\). The electric flux through a surface of area of 100 unit in \(X-Y\) plane is (a) 800 unit (b) 300 unit (c) 400 unit (d) 1500 unit
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