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 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) 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 1: 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.

Step 2: 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.

Step 3: 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:\[ E_B = \frac{-11.6 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} = -6.55 \times 10^5 \, \text{N/C} \]

Step 4: 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.

Step 5: 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.

Step 6: 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.

Key Concepts

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.