Problem 100 - University Physics with Modern Physics

Question

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 density 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 \(B\) .

Step-by-Step Solution

Answer

(a) 1187708.47 N/C to the right, (b) 1187708.47 N/C to the left, (c) 1187708.47 N/C to the right.

Step 1: Understand the Problem

The problem involves calculating the electric field created by two infinitely large charged sheets at specific points. The sheets have different uniform surface charge densities. The electric field due to an infinite sheet with surface charge density \( \sigma \) is given by \( E = \frac{\sigma}{2\varepsilon_0} \), where \( \varepsilon_0 \) is the permittivity of free space \( (8.85 \times 10^{-12} \, \mathrm{C}^{2}/\mathrm{N} \, \mathrm{m}^{2}) \). The direction of the electric field is away from the positively charged sheet and towards the negatively charged sheet.

Step 2: Calculate Electric Field Due to Each Sheet

Calculate the electric field due to each sheet A and B separately. For sheet A, \( \sigma = -9.50 \, \mu \mathrm{C} / \mathrm{m}^{2} \). Thus the electric field by sheet A is \( E_A = \frac{-9.50 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} \, \mathrm{N/C} \). For sheet B, \( \sigma = -11.6 \, \mu \mathrm{C} / \mathrm{m}^{2} \). Thus the electric field by sheet B is \( E_B = \frac{-11.6 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} \, \mathrm{N/C} \).

Step 3: Calculate Net Electric Field at Point (a)

Point (a) is 4.00 cm to the right of sheet A, inside the space between the sheets. Here, electric fields from both sheets contribute. The field from each negatively charged sheet points inward toward itself, so at this point, the fields add up: \( E_{\text{net}} = E_B - E_A \). Then compute \( E_{\text{net}} = \left| E_B \right| + \left| E_A \right| \) since their absolute values add up.

Step 4: Calculate Net Electric Field at Point (b)

Point (b) is 4.00 cm to the left of sheet A. The field due to sheet A points to the left (since it's negatively charged) and sheet B exerts a field to the left as well out of the sheets in this region, so: \( E_{\text{net}} = -E_A + (-E_B) = - \left( \left| E_A \right| + \left| E_B \right| \right) \).

Step 5: Calculate Net Electric Field at Point (c)

Point (c) is 4.00 cm to the right of sheet B. In this region, the electric field from sheet B points to the right and sheet A also contributes a field in the right direction since it is beyond sheet B, both fields add up: \( E_{\text{net}} = -E_A + (-E_B) = - \left( \left| E_A \right| + \left| E_B \right| \right) \).

Step 6: Final Calculations

Substitute the values to compute the net electric fields:- Step 3: \( E_{\text{net, (a)}} = \frac{11.6 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} + \frac{9.50 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} \) N/C.- Step 4: \( E_{\text{net, (b)}} = - \left[ \frac{9.50 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} + \frac{11.6 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} \right] \) N/C.- Step 5: The same expression as Step 4 but note its location as Step 5.

Key Concepts

Surface Charge Density

Surface charge density is a measure of how much electric charge is accumulated over a unit area on a surface. It is relevant when dealing with charged sheets, like Sheet A and Sheet B in our exercise. The surface charge density \( \sigma \) is expressed in coulombs per square meter (\( \mathrm{C/m}^2 \)).

In our scenario:

• Sheet A has a surface charge density of \(-9.50 \, \mu \mathrm{C/m}^2\).
• Sheet B has a surface charge density of \(-11.6 \, \mu \mathrm{C/m}^2\).

Both sheets have negative surface charge densities, which means the electric fields they produce point towards the respective sheets. Understanding surface charge density helps to determine the magnitude and direction of the electric field created by a uniformly charged sheet.

The electric field \( E \) due to a charged infinite sheet is directly proportional to its surface charge density and is calculated using the formula: \[ E = \frac{\sigma}{2\varepsilon_0} \] where \( \varepsilon_0 \) is the permittivity of free space.

Permittivity of Free Space

Permittivity of free space, denoted as \( \varepsilon_0 \), is a fundamental constant in electromagnetism. Its value is approximately \( 8.85 \times 10^{-12} \, \mathrm{C}^2/\mathrm{N} \, \mathrm{m}^2 \).

This constant measures the ability of a vacuum to permit electric field lines. Essentially, it quantifies the force between two point charges in vacuum conditions. In the context of electric fields around charged sheets, \( \varepsilon_0 \) is a crucial factor in calculating the electric field strength \( E \) produced by a charged sheet.

Thus, it appears in the formula:\[ E = \frac{\sigma}{2\varepsilon_0} \] This equation implies that an increase in permittivity reduces the strength of the electric field for a given surface charge density. Hence, the permittivity of free space is fundamental in assessing how charges interact in electrostatic situations.

Infinite Sheets

In physics problems involving electric fields, sheets are often treated as "infinite" to simplify calculations. When a sheet is assumed to be infinite, it suggests that its surface extends indefinitely in all directions with a uniform charge distribution.

This assumption allows us to conclude that the electric field it produces is uniform and perpendicular to the surface, remaining constant at each point in space around the sheet. There are several implications of considering sheets as infinite:

• The electric field is independent of the distance from the sheet; it does not weaken with distance.
• The edge effects are neglected because the sheet is considered to extend indefinitely.

In our problem, Sheets A and B are treated as infinite. This means the formula \( E = \frac{\sigma}{2\varepsilon_0} \) is valid at any distance from the sheet, providing a uniform field that simplifies the complexity of calculations.

Net Electric Field Calculation

The net electric field is the combined effect of electric fields produced by multiple sources. It can be understood as the vector sum of individual electric fields.

In our exercise, the electric fields from Sheets A and B are summed to find the net field at particular points. It's essential to consider the direction of the fields:

• Each negative charge in Sheets A and B causes electric fields pointing towards each respective sheet.
• To find the net field, one adds the magnitudes of both fields while considering their directions.

For example, at a point 4.00 cm right of Sheet A (between the sheets), the net field \( E_{\text{net}} \) is computed as:\[ E_{\text{net}} = |E_B| - |E_A| \].

For points outside the region between sheets (i.e., left of A and right of B), both fields add up as they point in the same direction:\[ E_{\text{net}} = -( |E_A| + |E_B| ) \].

Thus, calculating the net electric field involves understanding how individual fields interact based on their directions and magnitudes.