How is an electric field inversely proportional to the area of a cross section?
2 Answers
Actually, an electric field is not inversely proportional to the area of a cross section. In electrostatics, the electric field \(\mathbf{E}\) is related to the electric flux \(\Phi_E\) and the area \(A\) of the cross section by the equation:
\[ \Phi_E = \mathbf{E} \cdot A \]
where \(\Phi_E\) is the electric flux through the area \(A\).
The electric field \(\mathbf{E}\) itself is defined as the force \(\mathbf{F}\) per unit charge \(q\):
\[ \mathbf{E} = \frac{\mathbf{F}}{q} \]
In the context of a parallel plate capacitor, the electric field between the plates is given by:
\[ \mathbf{E} = \frac{V}{d} \]
where \(V\) is the potential difference between the plates and \(d\) is the separation between them. Here, the electric field \(\mathbf{E}\) is independent of the area of the plates.
So, the electric field is not inversely proportional to the area of the cross section. Instead, it remains constant for a given potential difference and distance between the plates. The area of the plates affects the capacitance of the capacitor, not the electric field directly.
\[ \Phi_E = \mathbf{E} \cdot A \]
where \(\Phi_E\) is the electric flux through the area \(A\).
The electric field \(\mathbf{E}\) itself is defined as the force \(\mathbf{F}\) per unit charge \(q\):
\[ \mathbf{E} = \frac{\mathbf{F}}{q} \]
In the context of a parallel plate capacitor, the electric field between the plates is given by:
\[ \mathbf{E} = \frac{V}{d} \]
where \(V\) is the potential difference between the plates and \(d\) is the separation between them. Here, the electric field \(\mathbf{E}\) is independent of the area of the plates.
So, the electric field is not inversely proportional to the area of the cross section. Instead, it remains constant for a given potential difference and distance between the plates. The area of the plates affects the capacitance of the capacitor, not the electric field directly.
The electric field \( E \) is related to the electric flux \( \Phi_E \) through a surface area \( A \) by the equation:
\[
\Phi_E = E \cdot A
\]
Where:
- \( \Phi_E \) is the electric flux,
- \( E \) is the electric field strength,
- \( A \) is the area of the cross section through which the electric field lines pass.
If the electric flux is constant (for example, in a closed surface), then we can rearrange this equation to express the electric field as:
\[
E = \frac{\Phi_E}{A}
\]
From this equation, it's clear that if the area \( A \) increases while keeping the electric flux \( \Phi_E \) constant, the electric field \( E \) must decrease. This means that the electric field is inversely proportional to the area of the cross section.
In summary, if the cross-sectional area increases, the electric field strength decreases, assuming the electric flux remains unchanged.
\[
\Phi_E = E \cdot A
\]
Where:
- \( \Phi_E \) is the electric flux,
- \( E \) is the electric field strength,
- \( A \) is the area of the cross section through which the electric field lines pass.
If the electric flux is constant (for example, in a closed surface), then we can rearrange this equation to express the electric field as:
\[
E = \frac{\Phi_E}{A}
\]
From this equation, it's clear that if the area \( A \) increases while keeping the electric flux \( \Phi_E \) constant, the electric field \( E \) must decrease. This means that the electric field is inversely proportional to the area of the cross section.
In summary, if the cross-sectional area increases, the electric field strength decreases, assuming the electric flux remains unchanged.