How is an electric field inversely proportional to the area of A cross section?
2 Answers
The electric field (denoted as **E**) is not directly related to the cross-sectional area (**A**) of a conductor in all cases. However, the inverse proportionality you are asking about could arise in a few contexts. Let me explain the key ideas that might help you understand this concept.
### 1. **In the context of a charged conductor (Gauss's Law):**
For a conductor, the electric field outside the surface can be understood using **Gauss's Law**, which relates the electric flux through a surface to the charge enclosed by that surface. In simplified form, Gauss's Law is:
\[
\Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0}
\]
where:
- \(\Phi_E\) is the electric flux,
- \(Q_{\text{enc}}\) is the enclosed charge, and
- \(\epsilon_0\) is the permittivity of free space.
For a conductor, the electric flux through a surface area \(A\) is related to the electric field **E**:
\[
\Phi_E = E \cdot A
\]
If the enclosed charge remains constant, an increase in the cross-sectional area \(A\) will result in a decrease in the electric field, assuming symmetry. Thus, the electric field becomes **inversely proportional to the area**:
\[
E \propto \frac{1}{A}
\]
### 2. **In the context of current density (J = I/A):**
In the case of current flow through a conductor, the **current density** (\(J\)) is related to the current (\(I\)) and the cross-sectional area \(A\) as:
\[
J = \frac{I}{A}
\]
The electric field **E** is related to the current density by Ohm's Law (in a material with conductivity \(\sigma\)):
\[
E = \frac{J}{\sigma}
\]
Substituting for \(J\):
\[
E = \frac{I}{\sigma A}
\]
In this case, the electric field **E** is inversely proportional to the cross-sectional area \(A\), meaning if the cross-sectional area increases, the electric field decreases.
### 3. **In a Capacitor (Parallel Plate):**
For a parallel plate capacitor, the electric field between the plates is:
\[
E = \frac{\sigma}{\epsilon_0}
\]
where:
- \(\sigma = \frac{Q}{A}\) is the surface charge density,
- \(Q\) is the charge on the plates, and
- \(A\) is the area of the plates.
Here, \(E = \frac{Q}{\epsilon_0 A}\), which shows that for a given charge, the electric field **E** is inversely proportional to the cross-sectional area of the plates.
---
### Summary:
The electric field can be inversely proportional to the cross-sectional area in several contexts:
1. **Gauss's Law**: In regions around a charged object, larger surface areas decrease the electric field.
2. **Current Flow in a Conductor**: For a fixed current, increasing the cross-sectional area decreases the electric field due to reduced current density.
3. **Capacitors**: In a parallel plate capacitor, the electric field is inversely proportional to the plate's area.
The context you're referring to determines the exact relationship, but these are common cases where electric field and cross-sectional area show inverse proportionality.
### 1. **In the context of a charged conductor (Gauss's Law):**
For a conductor, the electric field outside the surface can be understood using **Gauss's Law**, which relates the electric flux through a surface to the charge enclosed by that surface. In simplified form, Gauss's Law is:
\[
\Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0}
\]
where:
- \(\Phi_E\) is the electric flux,
- \(Q_{\text{enc}}\) is the enclosed charge, and
- \(\epsilon_0\) is the permittivity of free space.
For a conductor, the electric flux through a surface area \(A\) is related to the electric field **E**:
\[
\Phi_E = E \cdot A
\]
If the enclosed charge remains constant, an increase in the cross-sectional area \(A\) will result in a decrease in the electric field, assuming symmetry. Thus, the electric field becomes **inversely proportional to the area**:
\[
E \propto \frac{1}{A}
\]
### 2. **In the context of current density (J = I/A):**
In the case of current flow through a conductor, the **current density** (\(J\)) is related to the current (\(I\)) and the cross-sectional area \(A\) as:
\[
J = \frac{I}{A}
\]
The electric field **E** is related to the current density by Ohm's Law (in a material with conductivity \(\sigma\)):
\[
E = \frac{J}{\sigma}
\]
Substituting for \(J\):
\[
E = \frac{I}{\sigma A}
\]
In this case, the electric field **E** is inversely proportional to the cross-sectional area \(A\), meaning if the cross-sectional area increases, the electric field decreases.
### 3. **In a Capacitor (Parallel Plate):**
For a parallel plate capacitor, the electric field between the plates is:
\[
E = \frac{\sigma}{\epsilon_0}
\]
where:
- \(\sigma = \frac{Q}{A}\) is the surface charge density,
- \(Q\) is the charge on the plates, and
- \(A\) is the area of the plates.
Here, \(E = \frac{Q}{\epsilon_0 A}\), which shows that for a given charge, the electric field **E** is inversely proportional to the cross-sectional area of the plates.
---
### Summary:
The electric field can be inversely proportional to the cross-sectional area in several contexts:
1. **Gauss's Law**: In regions around a charged object, larger surface areas decrease the electric field.
2. **Current Flow in a Conductor**: For a fixed current, increasing the cross-sectional area decreases the electric field due to reduced current density.
3. **Capacitors**: In a parallel plate capacitor, the electric field is inversely proportional to the plate's area.
The context you're referring to determines the exact relationship, but these are common cases where electric field and cross-sectional area show inverse proportionality.
The relationship between an electric field and the area of a cross-section typically relates to the concept of electric flux. Electric flux (\(\Phi_E\)) through a surface is given by:
\[ \Phi_E = \mathbf{E} \cdot \mathbf{A} \]
where \(\mathbf{E}\) is the electric field vector, and \(\mathbf{A}\) is the area vector of the surface.
If the electric flux is constant and the area of the surface increases, the electric field must decrease to keep the flux constant. Conversely, if the area decreases, the electric field must increase. This is because:
\[ \mathbf{E} = \frac{\Phi_E}{\mathbf{A}} \]
So, if \(\Phi_E\) is constant and \(\mathbf{A}\) increases, \(\mathbf{E}\) decreases, and vice versa. Therefore, the electric field \(\mathbf{E}\) is inversely proportional to the area of the cross-section when the electric flux is constant.
\[ \Phi_E = \mathbf{E} \cdot \mathbf{A} \]
where \(\mathbf{E}\) is the electric field vector, and \(\mathbf{A}\) is the area vector of the surface.
If the electric flux is constant and the area of the surface increases, the electric field must decrease to keep the flux constant. Conversely, if the area decreases, the electric field must increase. This is because:
\[ \mathbf{E} = \frac{\Phi_E}{\mathbf{A}} \]
So, if \(\Phi_E\) is constant and \(\mathbf{A}\) increases, \(\mathbf{E}\) decreases, and vice versa. Therefore, the electric field \(\mathbf{E}\) is inversely proportional to the area of the cross-section when the electric flux is constant.