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 i, and appropriate constants.
Understand the Problem
The question is asking for a mathematical derivation of the electric field along the x-axis due to an infinitely long sheet of charge. It specifies that the derivation should be expressed in terms of given variables and constants.
Answer
The electric field along the x-axis for points outside the sheet ($x > \frac{L}{2}$) is given by: $$ \vec{E} = \frac{\eta}{\epsilon_0} \hat{i} $$
Answer for screen readers
The expression for the electric field along the x-axis for points outside the sheet ($x > \frac{L}{2}$) is: $$ \vec{E} = \frac{\eta}{\epsilon_0} \hat{i} $$
Steps to Solve
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Understanding Charge Distribution We know that the electric field produced by an infinitely long sheet of charge with surface charge density $\eta$ is constant and is given by the formula: $$ E = \frac{\eta}{2 \epsilon_0} $$
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Electric Field Direction Since the sheet lies in the xy-plane and we are looking for the electric field along the x-axis at points where $x > \frac{L}{2}$, the electric field is directed away from the sheet for positive charge density $\eta$. Thus, the direction of $\vec{E}$ will be in the positive $\hat{i}$ direction.
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Calculating the Total Electric Field For points outside the sheet at $x > \frac{L}{2}$, the electric field due to the sheet can be thought to arise from two equal contributions:
- Each side of the sheet contributes: $$ E = \frac{\eta}{2 \epsilon_0} $$
Thus, for points outside the sheet: $$ E_{\text{total}} = \frac{\eta}{2 \epsilon_0} + \frac{\eta}{2 \epsilon_0} = \frac{\eta}{\epsilon_0} $$
- Final Expression for the Electric Field The electric field at a point along the x-axis where $x > \frac{L}{2}$ can thus be expressed as: $$ \vec{E} = \frac{\eta}{\epsilon_0} \hat{i} $$
The expression for the electric field along the x-axis for points outside the sheet ($x > \frac{L}{2}$) is: $$ \vec{E} = \frac{\eta}{\epsilon_0} \hat{i} $$
More Information
The derived expression shows that the electric field due to an infinitely long sheet of charge is uniform and depends only on the surface charge density $\eta$, not on the distance from the sheet. This consistency is characteristic of infinite charge distributions.
Tips
- A common mistake is to forget that the electric field produced by the sheet is in both directions and must be summed appropriately.
- Misinterpreting the direction of the electric field; always verify that it points away from positive charge and toward negative charge.
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