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Two very large parallel sheets are 5.00 \(\mathrm{cm}\) apart. Sheet \(A\) carries a uniform surface charge density of \(-9.50 \mu \mathrm{C} / \mathrm{m}^{2}\) , and sheet \(B\) , which is to the right of \(A,\) carries a uniform charge of \(-11.6 \mu \mathrm{C} / \mathrm{m}^{2} .\) Assume the sheets are large enough to be treated as infinite. Find the magnitude and direction of the net electric field these sheets produce at a point (a) 4.00 \(\mathrm{cm}\) to the right of sheet \(A\) ; (b) 4.00 \(\mathrm{cm}\) to the left of sheet \(A ;(c) 4.00 \mathrm{cm}\) to the right of sheet \(B\) .

Short Answer

Expert verified
(a) 11.92 x 10^5 N/C to the left; (b) 1.18 x 10^5 N/C to the right; (c) 1.18 x 10^5 N/C to the left.

Step by step solution

01

Understand the problem setup

There are two infinite sheets with given surface charge densities separated by a distance of 5.00 cm. We need to find the electric field at specific points around these sheets. The charges on sheets A and B are negative, with sheet B having a greater magnitude of charge density than sheet A.
02

Electric field due to an infinite sheet

The electric field due to a single infinite plane sheet with surface charge density \(\sigma\) is \(E = \frac{\sigma}{2\varepsilon_0}\), where \(\varepsilon_0\) is the permittivity of free space \(\approx 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2\). The field is directed away from the sheet if the charge is positive and towards the sheet if the charge is negative.
03

Calculate electric fields due to each sheet

For sheet A with surface charge density \(\sigma_A = -9.50 \, \mu\text{C}/\text{m}^2\), we have:\[ E_A = \frac{-9.50 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} = -5.37 \times 10^5 \, \text{N/C} \]For sheet B with surface charge density \(\sigma_B = -11.6 \, \mu\text{C}/\text{m}^2\), we have:\[ E_B = \frac{-11.6 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} = -6.55 \times 10^5 \, \text{N/C} \]
04

Point a - 4.00 cm to the right of sheet A

At this point, we are between the two sheets. The electric field due to sheet A is directed towards the left, and the field due to sheet B is also towards the left. Thus, these fields add up:\[ E_{net_a} = |E_A| + |E_B| = 5.37 \times 10^5 + 6.55 \times 10^5 = 11.92 \times 10^5 \, \text{N/C} \]The direction is to the left.
05

Point b - 4.00 cm to the left of sheet A

Here we are left of both sheets. The field due to sheet A is to the left, and due to sheet B is to the right. Therefore, they subtract:\[ E_{net_b} = |E_B| - |E_A| = 6.55 \times 10^5 - 5.37 \times 10^5 = 1.18 \times 10^5 \, \text{N/C} \]The direction is to the right.
06

Point c - 4.00 cm to the right of sheet B

At this point, we are to the right of both sheets. The fields due to both sheets are directed towards the sheets, so they subtract:\[ E_{net_c} = |E_A| - |E_B| = 5.37 \times 10^5 - 6.55 \times 10^5 = 1.18 \times 10^5 \, \text{N/C} \]The direction is to the left.

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Key Concepts

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

Surface Charge Density
Surface charge density is a key concept when dealing with electric fields, especially in the context of parallel plates or sheets. It is denoted by the symbol \( \sigma \) and represents the amount of charge per unit area on a surface. For this exercise, sheet \( A \) has a surface charge density of \(-9.50 \, \mu C/m^2\), and sheet \( B \) has \(-11.6 \, \mu C/m^2\).

Understanding surface charge density helps in determining the electric field generated by a charged surface. With higher surface charge densities, the electric field produced will be stronger. This is because more charges exert greater force in a given space. Thus, calculating the surface charge density allows us to predict the intensity of the electric field effectively.

Moreover, surface charge density can indicate the polarity and sign of the electric field direction. Negatively charged surfaces imply that the electric field vectors are directed towards the surface, whereas, for positively charged surfaces, the vectors are directed away.
Permittivity of Free Space
The permittivity of free space, symbolized as \( \varepsilon_0 \), is a fundamental constant in electromagnetism. It is approximately equal to \( 8.85 \times 10^{-12} \, C^2/N \cdot m^2 \).

This constant is crucial in calculating electric fields because it relates the electric field to the surface charge density through the formula \( E = \frac{\sigma}{2\varepsilon_0} \) for an infinite plane sheet. Essentially, \( \varepsilon_0 \) quantifies how much electric field (flux) is permitted to "flow" through a vacuum or free space in response to a given charge. A lower permittivity would mean a stronger field generated for the same charge density.

When solving problems involving electrical forces, especially between charged objects such as plates or sheets, the permittivity of free space helps standardize the comparison and calculation of forces and fields, providing a uniform reference point in electromagnetics.
Electric Field Calculation
Electric field calculation is a systematic way of determining the magnitude and direction of the electric field from a charged object. For an infinite sheet of charge, this is given by the formula \( E = \frac{\sigma}{2\varepsilon_0} \).

In this problem, the electric field from each sheet is calculated first before determining the net electric field at a point. For sheet \( A \), with a surface charge density of \(-9.50 \, \mu\text{C}/\text{m}^2\), the electric field \( E_A \) is computed as follows:
  • Calculate using \( E_A = \frac{-9.50 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} = -5.37 \times 10^5 \, \text{N/C} \).
For sheet \( B \), with \(-11.6 \, \mu\text{C}/\text{m}^2\), the calculation is:
  • \( E_B = \frac{-11.6 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} = -6.55 \times 10^5 \, \text{N/C} \).
By calculating the electric fields separately, we then determine their net effect based on their respective directions, as the combination of these fields changes at different points in space.
Direction of Electric Fields
The direction of electric fields is fundamental to understanding how fields influence charges in their vicinity. For infinite charge sheets, the direction of the electric field is described as:
  • Away from the surface if positively charged.
  • Toward the surface if negatively charged.
In the given exercise, since both sheets \( A \) and \( B \) carry negative charges, the electric field from each is directed towards the respective sheet.

To determine the net electric field direction at specific points:
  • At 4.00 cm right of sheet \( A \), the fields from \( A \) and \( B \) both point to the left, leading to an addition of magnitudes.
  • At 4.00 cm left of sheet \( A \), the field from \( A \) points left while from \( B \), it points right, leading to a subtraction of magnitudes, with the net field to the right.
  • At 4.00 cm right of sheet \( B \), both field directions point left, resulting in subtracted magnitudes with the net field left.
Recognizing the direction in which fields act informs us about potential force interactions with other charges placed in these regions.

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Most popular questions from this chapter

(a) Sketch the electric field lines for an infinite line of charge. You may find it helpful to show the field lines in a plane containing the line of charge in one sketch and the field lines in a a plane perpendicular to the line of charge in a second sketch. (b) Explain how your sketches show (i) that the magnitude \(E\) of the electric field depends only on the distance \(r\) from the line of charge and (ii) that \(E\) decreases like \(1 / r .\)

A dipole consisting of charges \(\pm e, 220 \mathrm{nm}\) apart, is placed between two very large (essentially infinite) sheets carrying equal but opposite charge densities of 125\(\mu \mathrm{C} / \mathrm{m}^{2}\) (a) What is the maximum potential energy this dipole can have due to the sheets, and how should it be oriented relative to the sheets to attain this value? (b) What is the maximum torgue the sheets can exert on the dipole, and how should it be oriented relative to the sheets to attain this value? (c) What net force do the two sheets exert on the dipole?

Two small, copper spheres each have radius 1.00 \(\mathrm{mm}\) . (a) How many atoms does each sphere contain? (b) Assume that each copper atom contains 29 protons and 29 electrons. We know that electrons and protons have charges of exactly the same magnitude, but let's explore the effect of small differences (see also Problem 21.83\()\) . If the charge of a proton is \(+e\) and the magnitude of the charge of an electron is 0.100\(\%\) smaller, what is the net charge of each sphere and what force would one sphere exert on the other if they were separated by 1.00 \(\mathrm{m} ?\)

Lightning occurs when there is a flow of electric charge (principally electrons) between the ground and a thundercloud. The maximum rate of charge flow in a lightning bolt is about \(20,000 \mathrm{C} / \mathrm{s} ;\) this lasts for 100\(\mu \mathrm{s}\) or less. How much charge flows between the ground and the cloud in this time? How many electrons flow during this time?

Two very small \(8.55-\) g spheres, 15.0 \(\mathrm{cm}\) apart from center to center, are charged by adding equal numbers of electrons to each of them. Disregarding all other forces, how many electrons would you have to add to each sphere so that the two spheres will accelerate at 25.0 \(\mathrm{g}\) when released? Which way will they accelerate?

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