## **1) Intensity of a Field Inside a Conductor**
### **Introduction**
A conductor is a material consisting of neutral atoms, where [[charge]] resides on the surface rather than being distributed throughout the volume. This results in the [[Electric Field]] ($\mathbf{E}$) inside the conductor being zero.
### **Explanation**
To understand why the electric field is zero, consider two imaginary Gaussian surfaces:
- **$S_1$**: Located just beneath the conductor's surface.
- **$S_2$**: Positioned just outside the conductor's surface.
![[Pasted image 20241130095833.png]]
#### **a) When the Conductor is Neutral**
- **Charge Distribution**: The conductor contains no excess charge, so the enclosed charge ($q$) inside $S_1$ is zero.
- **No Surface Charge**: Since there is no surface charge, no flux passes through $S_2$.
Using **[[Gauss's Law]]**:
$
\phi_E = \frac{q}{\varepsilon_0}
$
Since $q = 0$, the electric field ($\mathbf{E}$) is:
$
E = 0
$
#### **b) When the Conductor Has Extra Charge**
- **Surface Distribution**: Any excess charge resides on the surface of the conductor, ensuring that $q$ enclosed by $S_1$ remains zero.
- **Flux Through $S_2$**: As charge resides on the surface, flux passes through $S_2$, resulting in an electric field only on the surface of the conductor.
### **Conclusion**
The electric field inside a conductor is always zero, regardless of whether it is neutral or carries excess charge.
### **Examples**
- **Electronics Shielding**: Sensitive circuits are placed in metallic enclosures to block external electric fields.
- **Airplane Safety**: Passengers remain safe during lightning strikes because the metallic shell of the airplane blocks electric fields.
---
## **2) Electric Intensity Due to an Infinite Sheet of Charge**
### **Arrangement**
- **Setup**: An infinite plane sheet has a uniform surface charge density ($\sigma$). Gaussian surfaces ($S_1$ and $S_2$) are positioned equidistant on either side.
- **Shape**: The Gaussian surface is a cylinder of cross-sectional area $A$, oriented perpendicularly to the sheet.
- **Field Properties**:
- Perpendicular to the flat faces ($\overrightarrow{\mathbf{E}} \parallel \overrightarrow{\mathbf{A}}$, $\theta = 0^\circ$).
- Parallel to the curved surface ($\overrightarrow{\mathbf{E}} \perp \overrightarrow{\mathbf{A}}$, $\theta = 90^\circ$).
![[Pasted image 20241130095848.png]]
### **Determination of [[Electric Field]] (E)**
1. **From [[Gauss's Law]]**:
$
\phi_E = \frac{q}{\varepsilon_0}
$
2. **Surface Charge Density**:
$
\sigma = \frac{q}{A} \implies q = \sigma A
$
Substituting $q = \sigma A$ into Gauss’s law:
$
\phi_E = \frac{\sigma A}{\varepsilon_0}
$
#### Flux Calculations:
- **Right End Flat Surface ($S_1$)**:
$
\phi_{e1} = EA \cos 0^\circ = EA
$
- **Left End Flat Surface ($S_2$)**:
$
\phi_{e2} = EA \cos 0^\circ = EA
$
- **Curved Surface**:
$
\phi_{e3} = EA \cos 90^\circ = 0
$
#### Total [[Electric Flux]]:
$
\phi_E = \phi_{e1} + \phi_{e2} + \phi_{e3} = EA + EA + 0 = 2EA
$
By comparing:
$
2EA = \frac{\sigma A}{\varepsilon_0} \implies E = \frac{\sigma}{2\varepsilon_0}
$
### **Conclusion**
The electric field due to an infinite sheet of charge is uniform and directed perpendicular to the sheet. In vector form:
$
\vec{E} = \frac{\sigma}{2\varepsilon_0} \hat{r}
$
---
## **3) Electric Field Intensity Between Two Oppositely Charged Parallel Plates**
### **Arrangement**
- Two parallel plates with uniform surface charge densities:
- Positive plate: $+\sigma$.
- Negative plate: $-\sigma$.
- Plates are assumed to be infinite to ensure uniform fields with no edge effects.
![[Pasted image 20241130095916.png]]
### **Calculation of E**
1. **From [[Gauss's Law]]**:
$
\phi_E = \frac{q}{\varepsilon_0}
$
Substituting $q = \sigma A$:
$
\phi_E = \frac{\sigma A}{\varepsilon_0}
$
#### Flux Contributions:
- **Through the Box’s Sides**: No flux passes through the sides since $\mathbf{E} \perp \mathbf{A}$:
$
\phi_e = 0
$
- **Through the Top**: No flux, as the field inside the metal plate is zero:
$
\phi_e = 0
$
- **Through the Bottom**: Field lines pass through the bottom surface:
$
\phi_e = EA
$
#### Total Flux:
$
\phi_E = EA
$
By comparing:
$
EA = \frac{\sigma A}{\varepsilon_0} \implies E = \frac{\sigma}{\varepsilon_0}
$
### **Conclusion**
The electric field between oppositely charged plates is:
- **Uniform**: The field strength does not vary.
- **Perpendicular**: Normal to the plate surfaces.
- **Independent**: Unaffected by the distance between the plates.
In vector form:
$
\vec{E} = \frac{\sigma}{\varepsilon_0} \hat{r}
$
---
## **Summary**
| **Topic** | **Key Equation** | **Significance** |
| ------------------------ | ----------------------------------- | ------------------------------------------------------------------------------ |
| Field inside a conductor | $E = 0$ | Explains shielding in metal enclosures. |
| Infinite sheet of charge | $E = \frac{\sigma}{2\varepsilon_0}$ | Demonstrates uniform fields for infinite charge distributions. |
| Parallel plates | $E = \frac{\sigma}{\varepsilon_0}$ | Provides the basis for capacitors and field uniformity in engineering devices. |
### Real-World Applications
- **Conductors**: Shielding electronics from external fields.
- **Infinite Sheet**: Basis for designing uniform electric fields in experiments.
- **Parallel Plates**: Capacitor design and electrostatic applications.
---
## References
