Problem 52 - Master Resource Book in JEE Main Physics

Question

The gravitational field due to a mass distribution is \(I=k / x^{3}\) in the \(x\)-direction ( \(k\) is a constant). Taking the gravitational potential to be zero at infinity, its value at a distance \(x / \sqrt{2}\) is (a) \(k / x\) (b) \(k / 2 x\) (c) \(k / x^{2}\) (d) \(k / 2 x^{2}\)

Step-by-Step Solution

Answer

Option (d) \( \frac{k}{2x^2} \).

Step 1: Define Gravitational Potential

Gravitational potential is related to the gravitational field by the formula \( V(x) = - \int E(x) \, dx \), where \(E(x)\) is the gravitational field.

Step 2: Set Up the Integral

Given the gravitational field \( E(x) = \frac{k}{x^3} \), we need to integrate it to find the potential: \( V(x) = - \int \frac{k}{x^3} \, dx \).

Step 3: Integrate to Find Potential Function

Solving the integral \(- \int \frac{k}{x^3} \, dx\), we get \( V(x) = \frac{k}{2x^2} + C \).

Step 4: Apply Boundary Condition

Since the potential is zero at infinity \( V(\infty) = 0 \), this makes \(C=0\), resulting in \( V(x) = \frac{k}{2x^2} \).

Step 5: Calculate Potential at Given Distance

Substitute \(x/\sqrt{2}\) into the potential \(V(x)\) to find \(V\left(\frac{x}{\sqrt{2}}\right) = \frac{k}{2(\frac{x}{\sqrt{2}})^2} = \frac{k}{\frac{x^2}{2}} = \frac{k}{x^2/2} = \frac{k}{2x^2} \).

Step 6: Choose the Correct Option

The calculated potential \( \frac{k}{2x^2} \) corresponds to choice (d).

Key Concepts

Gravitational Field

The idea of a gravitational field is central to understanding how objects interact in space due to gravity. A gravitational field is essentially a region of space around a mass where a force is exerted on any other mass brought into the field.
The concept is important because it allows us to describe the gravitational influence of a body systematically. The strength of this field at a point is represented by the gravitational field value, often denoted as \( E(x) \), and can be thought of as the force per unit mass experienced by a small test mass placed at a particular location relative to the source mass.

• In the provided problem, we have a gravitational field given by the formula \( E(x) = \frac{k}{x^3} \), which implies that the field's strength decreases dramatically as we move away from the source along the \( x \)-direction.
• This negative exponent tells us that the force decreases with the cube of the distance \( x \), highlighting how rapidly gravitational influences diminish over space.

Understanding the gravitational field helps in comprehending how gravitational potential and integration play roles in quantifying energy and force distributions.

Integration in Physics

In physics, integration is an invaluable mathematical tool that helps to accumulate quantities over a continuous range. When dealing with fields like gravitational fields, integration allows us to transition from the local, differential quantity **(**the field strength, for example**)** to a global, cumulative quantity **(**such as potential energy**)**.

• The integral of a function gives us the area under the curve of that function over a specified range. In this exercise, integrating the gravitational field \( E(x) = \frac{k}{x^3} \) with respect to \( x \) involves finding the antiderivative that will yield the gravitational potential \( V(x) \).
• The formula for gravitational potential is given as \( V(x) = - \int E(x) \, dx \). Solving this integral involves reversing the process of differentiation: determining which function would differentiate to give \( \frac{k}{x^3} \).
• Upon integration, the potential function resulting from the exercise is \( V(x) = \frac{k}{2x^2} + C \), where \( C \) is an integration constant determined by boundary conditions.

The integral acts like a bridge, converting the force distribution across space into information about potential energy that can be used in practical calculations.

Boundary Conditions

Boundary conditions are the additional information needed to uniquely determine the solution to a differential equation problem. They ensure our physical insights align with reality by using known attributes of the system to fix unknown constants.

• In the context of gravitational potential, knowing that the potential is zero at infinity helps us to find the precise form of our potential function.
• For the example problem, the condition \( V(\infty) = 0 \) determines that the integration constant \( C = 0 \).
• Applying this condition leads to the finalized expression for the potential as \( V(x) = \frac{k}{2x^2} \).

Boundary conditions are crucial because they ground the mathematical solutions in physical reality, transforming general solutions into specific ones that can describe actual physical systems and scenarios.