Two parallel large charged sheets with sigma and 3sigma respectively. Find electric field on regions 1, 2, and 3.

Understand the Problem

The question describes two parallel, large charged sheets with surface charge densities (\sigma) and (3\sigma). It asks to determine the electric field in the three distinct regions created by these sheets: region 1 (to the left of both sheets), region 2 (between the sheets), and region 3 (to the right of both sheets). This involves applying Gauss's law and the principle of superposition to find the net electric field in each region due to the charged sheets.

Answer

Region 1: $E = \frac{2\sigma}{\epsilon_0}$ (left) Region 2: $E = \frac{\sigma}{\epsilon_0}$ (left) Region 3: $E = \frac{2\sigma}{\epsilon_0}$ (right)

Steps to Solve

Define the direction of the electric field due to each sheet

Since both sheets have positive surface charge densities, the electric field due to each sheet will point away from the sheet in both directions.

Calculate the electric field due to the first sheet (with charge density $\sigma$)

The electric field due to an infinite charged sheet is given by $E = \frac{\sigma}{2\epsilon_0}$, where $\sigma$ is the surface charge density and $\epsilon_0$ is the permittivity of free space. Thus, the electric field due to the first sheet is $E_1 = \frac{\sigma}{2\epsilon_0}$.

Calculate the electric field due to the second sheet (with charge density $3\sigma$)

Similarly, the electric field due to the second sheet is $E_2 = \frac{3\sigma}{2\epsilon_0}$.

Determine the net electric field in region 1 (left of both sheets)

In region 1, both electric fields point to the left. Since we typically consider the positive direction to be to the right, both $E_1$ and $E_2$ will be negative. The net electric field is $E_{net1} = -E_1 - E_2 = -\frac{\sigma}{2\epsilon_0} - \frac{3\sigma}{2\epsilon_0} = -\frac{4\sigma}{2\epsilon_0} = -\frac{2\sigma}{\epsilon_0}$. The magnitude is $\frac{2\sigma}{\epsilon_0}$, and the direction is to the left.

Determine the net electric field in region 2 (between the sheets)

In region 2, the electric field due to the first sheet points to the right (positive direction), and the electric field due to the second sheet points to the left (negative direction). The net electric field is $E_{net2} = E_1 - E_2 = \frac{\sigma}{2\epsilon_0} - \frac{3\sigma}{2\epsilon_0} = -\frac{2\sigma}{2\epsilon_0} = -\frac{\sigma}{\epsilon_0}$. The magnitude is $\frac{\sigma}{\epsilon_0}$, and the direction is to the left.

Determine the net electric field in region 3 (right of both sheets)

In region 3, both electric fields point to the right (positive direction). The net electric field is $E_{net3} = E_1 + E_2 = \frac{\sigma}{2\epsilon_0} + \frac{3\sigma}{2\epsilon_0} = \frac{4\sigma}{2\epsilon_0} = \frac{2\sigma}{\epsilon_0}$. The magnitude is $\frac{2\sigma}{\epsilon_0}$, and the direction is to the right.

Region 1: $E = \frac{2\sigma}{\epsilon_0}$ (left) Region 2: $E = \frac{\sigma}{\epsilon_0}$ (left) Region 3: $E = \frac{2\sigma}{\epsilon_0}$ (right)

More Information

The electric field due to an infinite sheet of charge is uniform and does not depend on the distance from the sheet. The direction of the field is away from the sheet if the charge is positive and toward the sheet if the charge is negative. The principle of superposition allows us to add the electric fields due to multiple charges to find the net electric field.

Tips

A common mistake is to forget the direction of the electric field due to each sheet. It is important to consider the direction when summing the fields in each region. Also, not using absolute values can lead to errors if the directions are not carefully considered with positive and negative signs. Another mistake is to incorrectly apply the formula for the electric field of an infinite sheet, either by forgetting the factor of 2 in the denominator or by confusing $\sigma$ with the total charge.

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