/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 A large flat horizontal sheet of... [FREE SOLUTION] | 91Ó°ÊÓ

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A large flat horizontal sheet of charge has a charge per unit area of \(9.00 \mu \mathrm{C} / \mathrm{m}^{2} .\) Find the electric field just above the middle of the sheet.

Short Answer

Expert verified
The electric field just above the middle of the sheet is \(5.08 × 10^{5} N/C\).

Step by step solution

01

Identify given parameters

In this case, the given surface charge density, \( \sigma \), is \(9.00 \mu C/m^2\). Note that we will need to convert this to standard units, i.e., from \( \mu C/m^2 \) to \( C/m^2 \).
02

Convert units

We know that \(1 \mu C = 1×10^{-6} C\). Thus, the given surface charge density in standard units is \(9.00 × 10^{-6} C/m^2\).
03

Calculate the electric field

Now we can simply apply the formula for the electric field near a large, flat, charged sheet, \(E = \frac{\sigma}{2\epsilon_0}\). This gives us \(E = \frac{9.00 × 10^{-6} C/m^2}{2×8.85 × 10^{-12} C^2/(N m^2)}\). Dividing these numbers gives an electric field \(E = 5.08 × 10^{5} N/C\).

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Key Concepts

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

Surface Charge Density
When we talk about surface charge density, we are referring to the amount of electric charge present per unit area on a surface.
The symbol for surface charge density is \( \sigma \), and it's commonly expressed in units of coulombs per square meter (\( C/m^2 \)). In the problem, the surface charge density is given as \( 9.00 \mu C/m^2 \). Converting this to standard units, you would multiply by \( 1 \times 10^{-6} \), resulting in \( 9.00 \times 10^{-6} C/m^2 \).
This conversion is vital for accurate calculations.
Understanding surface charge density is crucial, as it helps determine the strength and direction of the electric field generated by charged surfaces. The higher the surface charge density, the stronger the electric field produced.
Gauss's Law
Gauss's Law is a fundamental principle in electrostatics, connecting electric fields to the distribution of electric charge.
It provides a powerful way to calculate electric fields when the symmetry of the problem allows. According to Gauss's Law, the electric flux through a closed surface is equal to the charge enclosed divided by the electric constant \( \epsilon_0 \), which is approximately \( 8.85 \times 10^{-12} C^2/(N \cdot m^2) \).
For a uniformly charged infinite plane, like the one described in the exercise, Gauss’s Law simplifies the calculation of the electric field significantly. Using a hypothetical Gaussian "pillbox" that intersects the plane, the symmetry allows us to find that the electric field just above this surface is \( E = \frac{\sigma}{2\epsilon_0} \).
This expression shows us that the electric field depends only on the surface charge density and the electric constant.
Electrostatics
Electrostatics is the branch of physics that studies stationary electric charges or charges at rest.
These charges produce electric fields and interact with each other, causing forces without any movement due to the static nature. One fundamental aspect of electrostatics is that like charges repel each other, while opposite charges attract.
This understanding helps explain how charged surfaces exert forces and create electric fields.
In electrostatics, problems often involve calculating the effect of an electric field due to charged objects. In our exercise, for instance, the electric field caused by a charged sheet is determined by the surface charge density, as the charges are static and distributed uniformly.
Factors such as the medium between charges and their separation influence the forces experienced due to the electric field.

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Most popular questions from this chapter

A uniformly charged, straight filament \(7.00 \mathrm{m}\) in length has a total positive charge of \(2.00 \mu \mathrm{C} .\) An uncharged cardboard cylinder \(2.00 \mathrm{cm}\) in length and \(10.0 \mathrm{cm}\) in radius surrounds the filament at its center, with the filament as the axis of the cylinder. Using reasonable approximations, find (a) the electric field at the surface of the cylinder and (b) the total electric flux through the cylinder.

A solid copper sphere of radius \(15.0 \mathrm{cm}\) carries a charge of \(40.0 \mathrm{nC} .\) Find the electric field (a) \(12.0 \mathrm{cm},\) (b) \(17.0 \mathrm{cm}\) and (c) \(75.0 \mathrm{cm}\) from the center of the sphere. (d) What If? How would your answers change if the sphere were hollow?

A conducting spherical shell of radius \(15.0 \mathrm{cm}\) carries a net charge of \(-6.40 \mu \mathrm{C}\) uniformly distributed on its surface. Find the electric field at points (a) just outside the shell and (b) inside the shell.

Review problem. An early (incorrect) model of the hydrogen atom, suggested by J.J. Thomson, proposed that a positive cloud of charge \(+e\) was uniformly distributed throughout the volume of a sphere of radius \(R,\) with the electron an equal-magnitude negative point charge \(-e\) at the center. (a) Using Gauss's law, show that the electron would be in equilibrium at the center and, if displaced from the center a distance \(r

Consider two identical conducting spheres whose surfaces are separated by a small distance. One sphere is given a large net positive charge while the other is given a small net positive charge. It is found that the force between them is attractive even though both spheres have net charges of the same sign. Explain how this is possible.

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