Problem 177 - A Complete Resource Book in Physics for JEE Main

Question

Assume that an electric field \(\vec{E}=30 x^{2} \hat{i}\) exists in space. Then the potential difference \(V_{A}-V_{O}\), where \(V_{O}\) is the potential at the origin and \(V_{A}\) the potential at \(x=2 \mathrm{~m}\) is (A) \(120 \mathrm{~J}\) (B) \(-120 \mathrm{~J}\) (C) \(-80 \mathrm{j}\) (D) \(80 \mathrm{~J}\)

Step-by-Step Solution

Answer

The potential difference between points A and O in the given electric field \(\vec{E} = 30x^2 \hat{i}\) can be calculated using the formula: \(V_{A} - V_{O} = -\int_{C} \vec{E} \cdot \vec{dl}\). After setting up the integral, integrating the electric field function, and evaluating the integral, we find that \(V_{A} - V_{O} = -80\, J\). Therefore, the correct answer is (C) \(-80\, J\).

Step 1: Understand the electric field and potential difference formula

For an electric field \(\vec{E}\), the potential difference between any two points can be found using the formula: \[V_{A} - V_{O} = -\int_{C} \vec{E} \cdot \vec{dl}\] The integral of \(\vec{E} \cdot \vec{dl}\) will be calculated along a path(C) from point O (at the origin) to point A (at \(x = 2\,m\)). In this exercise, the electric field is given as \(\vec{E} = 30x^2 \hat{i}\).

Step 2: Set up the integral

Since the electric field is only in the x-direction and we are only considering the x-component of the path, the integral becomes: \[\int_{0}^{2} (-30x^2) \, dx\] We'll integrate with respect to x from 0 to 2.

Step 3: Integrate the electric field function

Integrate the function with respect to x: \[ -\int_0^2 (30x^2) dx = -10x^3\Big|_0^2\]

Step 4: Evaluate the integral

Plug in the limits of the integration to find the potential difference: \[V_{A} - V_{O} = -10(2^3) + 10(0^3) = -80\] The potential difference \(V_{A} - V_{O} = -80\, J\). Thus, the correct answer is (C) \(-80\, J\).

Key Concepts

Electric Field Calculation

When dealing with electric fields, the first step is to identify the characteristics of the field itself. In this problem, the electric field \(\vec{E}\) is defined as \(30 x^2 \hat{i}\), which means that it is a vector field oriented solely along the x-direction, and its magnitude changes with the square of the position \(x\).
This type of electric field is non-uniform because its strength increases as you move further along the x-axis.
Understanding the configuration of the electric field is crucial for any subsequent calculations. Here are some key insights about electric fields:

• The electric field \(\vec{E}\) is often denoted in terms of its components; in this instance, it has an x-component only.
• It is expressed in units of newtons per coulomb (N/C) or volts per meter (V/m), highlighting the relation between force and electric charge.
• Knowing the field expression \(30 x^2\), we can anticipate that the field's strength increases as x increases, impacting the potential difference further.

Potential Difference Formula

Potential difference is a significant quantity in electromagnetism, representing the work done per unit charge in moving a test charge from one point to another in an electric field.
The formula to calculate the potential difference \(V_A - V_O\) is given by\[V_{A} - V_{O} = -\int_{C} \vec{E} \cdot \vec{dl}\]
This formula reveals that the potential difference is the negative line integral of the electric field over the path between two points.
Here's how the formula breaks down:

• The negative sign indicates that moving against the direction of an electric field requires work.
• \(\vec{dl}\) represents a small segment along the path, indicating the infinitesimal distance during the integration.
• In many scenarios, particularly when dealing with simple paths, the integration simplifies, as in moving straight along an axis.

The potential difference formula embodies one of the most direct connections between fields and potentials, allowing one to translate the abstract concept of a field into a more tangible understanding of potential.

Integration in Physics

In physics, integration is a powerful mathematical tool often used to find quantities like area, volume, displacement, and in this case, potential difference from given field expressions.
Integration enables us to handle continuous expressions and sum their contributions across intervals.
Here, the integration of the electric field \(30x^2\) over the specific x-bound from 0 to 2 is needed:

• Set up the definite integral: \(\int_{0}^{2} (-30x^2) \, dx\).
• Perform the integration: solve \(-\int_0^2 (30x^2) \, dx = -10x^3\Big|_0^2\) which simplifies to \(-10(2^3) + 10(0^3)\).
• By evaluating the result of the integration, we find the potential difference to be \(-80 \text{ J}\), explaining the negative sign as a decrease in potential energy moving in the field's direction.

Through this example, we see how integration connects vector fields with scalar potentials, providing a systematic method to derive potential differences across varying field intensities.