Answer :
Answer:The magnitude of the electric field intensity \( E \) varies along the surface of a pear-shaped charged conductor due to the differences in curvature at different points on the surface. This variation can be explained by the concept of surface charge density and the distribution of charge on conductors.
Here's a detailed explanation:
1. **Surface Charge Density**: The electric field just outside a charged conductor is directly proportional to the surface charge density (σ) at that point. Mathematically, \( E = \frac{\sigma}{\epsilon_0} \), where \( \epsilon_0 \) is the permittivity of free space.
2. **Curvature of the Surface**: The surface charge density itself depends on the curvature of the surface. At points where the curvature is high (such as the narrow ends or tips), the surface charge density is higher. Conversely, at points where the curvature is low (such as the broader parts of the pear shape), the surface charge density is lower.
3. **Distribution of Charges**: On a conductor, charges repel each other and tend to distribute themselves in such a way that the electric field inside the conductor is zero. This distribution leads to more charges accumulating at regions of higher curvature.
4. **Resulting Electric Field**: As a result of the higher surface charge density at points of higher curvature, the electric field intensity \( E \) is greater at these points. At points with lower curvature, the electric field intensity is correspondingly lower.
So, in summary, the electric field intensity \( E \) varies along the surface of a pear-shaped charged conductor because the surface charge density varies with the curvature of the surface. Higher curvature regions have higher charge density and, thus, a stronger electric field, while lower curvature regions have lower charge density and a weaker electric field.
Explanation:
