An infinitely long sheet of charge of width L lies in the xy-plane between x = -L/2 and x = L/2. The surface charge density is η. Derive an expression for the electric field E alon... An infinitely long sheet of charge of width L lies in the xy-plane between x = -L/2 and x = L/2. The surface charge density is η. Derive an expression for the electric field E along the x-axis for points outside the sheet (x > L/2) for x > 0. Express your answer in terms of the variables η, x, L, unit vector î, and appropriate constants.
Understand the Problem
The question asks to derive an expression for the electric field along the x-axis for points outside an infinitely long sheet of charge. It specifies the dimensions of the sheet and the charge density, indicating that the solution should express the electric field in terms of given variables and constants.
Answer
The electric field along the x-axis for points outside the sheet is given by $\vec{E} = \frac{\eta}{2\epsilon_0} \hat{i}$.
Answer for screen readers
The expression for the electric field along the x-axis for points outside the sheet is:
$$ \vec{E} = \frac{\eta}{2\epsilon_0} \hat{i} $$
Steps to Solve
- Identify Parameters and Variables
Define the parameters:
- Surface charge density: $\eta$
- Width of the sheet: $L$
- Position away from the sheet: $x$ (where $x > \frac{L}{2}$)
- Electric Field Due to an Infinite Plane Sheet of Charge
For an infinite plane sheet with uniform surface charge density $\eta$, the electric field (E) at a distance $x$ from the sheet is given by the formula:
$$ E = \frac{\eta}{2\epsilon_0} $$
where $\epsilon_0$ is the permittivity of free space.
- Considering the Direction of the Electric Field
Since we are looking for the electric field to the right of the sheet (where $x > \frac{L}{2}$), the electric field will be directed away from the sheet. Therefore, the net electric field for points to the right of the finite sheet will be:
$$ \vec{E} = \frac{\eta}{2\epsilon_0} \hat{i} $$
- Final Expression
Since we are examining the electric field only for $x > \frac{L}{2}$, there is no additional contribution from any parts of the sheet in this region. Thus, the final expression for the electric field along the x-axis is:
$$ \vec{E} = \frac{\eta}{2\epsilon_0} \hat{i} $$
The expression for the electric field along the x-axis for points outside the sheet is:
$$ \vec{E} = \frac{\eta}{2\epsilon_0} \hat{i} $$
More Information
This expression indicates that the electric field produced by an infinitely long sheet of charge is constant and uniform in the region outside the sheet. This means that at any distance $x > \frac{L}{2}$, the electric field will retain the same magnitude and direction.
Tips
- Neglecting the Direction: Sometimes, students forget to account for the direction of the electric field. It should always point away from the positively charged sheet or toward a negatively charged one.
- Confusing Finite with Infinite Sheets: It’s crucial to remember that the formulas differ for finite and infinite sheets of charge. The expression derived applies strictly to an infinite sheet.
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