Problem 30 - A Complete Resource Book in Physics for JEE Main

Question

The electric \(\vec{E}\) is given by \(\vec{E}=a \hat{i}+b \hat{j}\) (where \(a\) and \(b\) is constant and \(\hat{i}, \hat{j}\) are unit vector along \(x\) and \(y\) axis, respectively), the flux passing through a square area of side \(l\) and parallel to \(y=z\) plane is (A) \(b l^{2}\) (B) \(a l^{2}\) (C) \(\sqrt{\left(a^{2}+b^{2}\right)} l^{2}\) (D) \(\sqrt{\left(a^{2}-b^{2}\right)} l^{2}\)

Step-by-Step Solution

Answer

The electric flux through the square area is equal to \(a l^2\). So, the correct answer is (B) \(a l^2\).

Step 1: Determine the normal vector to the square

Since the square is parallel to the \(y=z\) plane, the normal vector to the square will be parallel to the \(x\)-axis. Let \(\hat{n}\) represent the normal vector to the square, then \(\hat{n} = \hat{i}\).

Step 2: Calculate the dot product of the electric field and the normal vector

The electric field is given by \(\vec{E} = a \hat{i} + b \hat{j}\) and the normal vector is \(\hat{n} = \hat{i}\). We need to calculate the dot product \(\vec{E} \cdot \hat{n}\) to find the component of the electric field in the \(x\)-direction. Using the dot product formula, we get: \(\vec{E} \cdot \hat{n} = (a \hat{i} + b \hat{j}) \cdot \hat{i} = a (\hat{i} \cdot \hat{i}) + b (\hat{j} \cdot \hat{i}) = a\)

Step 3: Integrate the dot product over the square surface

The electric flux through the square is given by the integral of the \(x\)-component of the electric field over the square: \(\Phi = \iint_S \vec{E} \cdot d\vec{A}\) Since the electric field is constant (i.e., \(a\) and \(b\) are constants) and the square has a side length of \(l\), we can rewrite the integral as a simple multiplication: \(\Phi = a \int_0^l \int_0^l dA = a l^2\)

Step 4: Identify the correct answer

The electric flux through the square area is equal to \(a l^2\). So, the correct answer is (B) \(a l^2\).

Key Concepts

Electric Field

The concept of the electric field is crucial in understanding electric flux. An electric field \(\vec{E}\) can be thought of as a force field that surrounds electric charges. It describes the force that a positive test charge would experience at any point in space. The direction of the electric field vector is the direction in which a positive charge would move if placed in the field.
\(\)

• Mathematically, an electric field is often expressed in vector form. In the given exercise, it's written as \(\vec{E}=a \hat{i} + b \hat{j}\), which means it has components along the x-axis and y-axis.
• The unit vectors \hat{i}\ and \hat{j}\ point along the x and y directions, respectively, and \(a\) and \(b\) are constants that determine the field's strength in these directions.

Understanding the electric field's vector nature is essential because it helps explain how electric flux, the flow of electric fields through a surface, is calculated in the next steps.

Dot Product

The dot product is a fundamental operation in vector mathematics. It's used to find the component of one vector in the direction of another. In the exercise, the dot product helps determine how much of the electric field passes through a given surface.
\(\)
Think of the dot product \(\vec{E} \cdot \hat{n}\) as the projection of the electric field on the normal vector \(\hat{n}\) of the surface. Here are some essential points:

• Given two vectors \(\vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}\) and \(\vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k}\), the dot product is calculated as \(\vec{A} \cdot \vec{B} = A_xB_x + A_yB_y + A_zB_z\).
• If vectors are perpendicular, their dot product is zero, indicating no component in that direction.

For the exercise, \(\vec{E} \cdot \hat{n} = a\), since \(\hat{n} = \hat{i}\) eliminates all other components except the one along the x-axis.

Vector Calculus

Vector calculus is a branch of mathematics that studies vectors and their operations. It plays a key role when dealing with fields like electric fields. In particular, calculating flux requires integrating over a surface, which involves vector calculus techniques.
\(\)
Here's how it fits into the exercise:

• Electric flux \(\Phi\) is found by integrating the dot product of the electric field with the differential area vector over the surface \(S\).
• The surface integral \(\Phi = \iint_S \vec{E} \cdot d\vec{A}\) simplifies to straightforward multiplication because the electric field is constant, and the area of the square is \(l^2\).

Understanding these operations helps in grasping how vector fields interact with surfaces to produce a measurable flux.

Gauss's Law

Gauss's Law is one of the four Maxwell's equations, which are fundamental to electromagnetism. It relates the electric flux flowing out of a closed surface to the charge enclosed by that surface. While Gauss's Law often deals with closed surfaces, its basic principle aids in understanding flux through any surface, such as the square in the exercise.
\(\)
Key notes about Gauss's Law:

• The law states that the total electric flux \(\Phi\) across a closed surface is equal to the charge enclosed \(Q_{ ext{enc}}\) divided by the permittivity of free space \(\varepsilon_0\): \[\Phi = \frac{Q_{\text{enc}}}{\varepsilon_0}\]
• It simplifies solving problems where symmetry allows for easy determination of the electric field over the surface.

Though not directly used in the exercise, Gauss's Law underpins the concept of flux and its relation to electric fields, providing a broader understanding of electromagnetism.