Problem 99 - Master Resource Book in JEE Main Physics

Question

In a region of space having a unifrom electric field \(E\), a hemispherical bowl of radius \(r\) is placed. The electric flux \phi through the bowl is (a) \(2 \pi r E\) (b) \(4 \pi r^{2} E\) (c) \(2 \pi r^{2} E\) (d) \(\pi r^{2} E\)

Step-by-Step Solution

Answer

The correct option is (c) \(2 \pi r^{2} E\).

Step 1: Recall the Electric Flux Definition

Electric flux \(\Phi\) through a surface is given by the equation \(\Phi = \vec{E} \cdot \vec{A}\), where \(\vec{E}\) is the electric field and \(\vec{A}\) is the area vector of the surface. For uniform \(\vec{E}\), this becomes \(\Phi = EA\).

Step 2: Consider the Surface Area of the Hemisphere

The surface area of a hemisphere is \(2\pi r^2\). The area vector \(\vec{A}\) is perpendicular to the surface, and for a hemisphere aligned with a uniform electric field, the component of \(\vec{A}\) that contributes to flux is perpendicular to the flat plane of the hemisphere.

Step 3: Use Gauss's Law for Calculation

Because the hemisphere is symmetry focused, the flux through the hemisphere's curved surface is half of the flux through a full sphere. The field lines exit through the flat part of the hemisphere. The flux through the curved surface can be modeled as half that of a full sphere, \(\Phi_{total} = (1/2) \cdot (4 \pi r^2) \cdot E = 2 \pi r^2 E\).

Key Concepts

Electric Field

The electric field, often symbolized by \( \vec{E} \), is a vector field around charged particles. It defines the force a positive test charge would experience at any point in space. The electric field is a crucial concept in electromagnetism and is measured in units of newtons per coulomb (N/C) or volts per meter (V/m). When dealing with a uniform electric field, it means that the strength and direction of the field are consistent over the space it occupies.
One can imagine electric field lines extending outwards from a positive charge, pointing towards a negative charge. The density of these lines correlates with the field's strength. In contexts like this exercise, understanding the direction and magnitude is essential for calculating phenomena such as electric flux. Especially in a uniformly distributed field, the assumptions become simpler, as only the alignment and area impact the calculus of the flux.

Gauss's Law

Gauss's Law is a fundamental principle connecting electric fields to the charges that generate them. It states that the total electric flux through a closed surface is equal to the charge enclosed divided by the electric constant \( \epsilon_0 \). Mathematically, it's expressed as: \[\Phi = \frac{Q_{enc}}{\epsilon_0}\]Where \( \Phi \) is the electric flux, and \( Q_{enc} \) is the enclosed charge.
Gauss's Law is especially effective in problems with high symmetry like spheres, cylinders, or regions with uniform electric fields. In this case, the hemisphere aligns within a uniform field. This symmetry simplifies the calculations, as the intuition from Gauss's Law tells us how the field interacts with the surface. For a closed symmetrical shape like a sphere or hemisphere, Gauss's Law helps simplify the interaction between the field and the surface, minimizing the complexity when calculating flux through either part of the structure.

Surface Area of Hemisphere

The surface area of a hemisphere is an essential metric for calculating electric flux through it. A hemisphere is half of a sphere, so its surface area is half that of a full sphere, which is \( 4\pi r^2 \). Hence, the surface area of a hemisphere can be calculated as \( 2\pi r^2 \).
In problems involving electric fields, we consider both the curved surface area and any planar bases (if applicable). For a hemisphere, it involves understanding how the electric field interacts with the larger curved surface. In this exercise, while dealing with a uniform electric field, only the curved surface substantially contributes to the flux because the field lines exit through the flat plane. Therefore, understanding how to compute and visualize the surface area is vital to determine the field's effect when integrated over a hemisphere.