Problem 40 - A Complete Resource Book in Physics for JEE Main

Question

If electric field is given by \(\vec{E}=\left(\frac{1}{x^{2}}\right) \hat{i} \mathrm{~V} / \mathrm{m}\), the magnitude of potential difference between points \(x=10 \mathrm{~cm}\) and \(x=20 \mathrm{~cm}\) is (A) \(\mathbb{V}\) (B) \(2 \mathrm{~V}\) (C) \(5 \mathrm{~V}\) (D) \(10 \mathrm{~V}\)

Step-by-Step Solution

Answer

The magnitude of the potential difference between the points \(x=10 \thinspace cm\) and \(x=20 \thinspace cm\) is (C) \(5 \mathrm{~V}\).

Step 1: Expression for the Electric Field in terms of Potential Difference

The electric field \(\vec{E}\) can be related to the potential difference \(V\) through the expression \[V = -\int_{x_1}^{x_2} \vec{E} \cdot \mathrm{d}\vec{r}\] Where \(x_1 = 10 cm\) and \(x_2 = 20 cm\). We are given \(\vec{E} = \left(\frac{1}{x^2}\right) \hat{i} \, \frac{V}{m}\).

Step 2: Integrate Electric Field along the Path

Now we integrate \(\vec{E}\) along the path (\(x\)-axis) from \(x_1\) to \(x_2\): \[V = -\int_{x_1}^{x_2} \left(\frac{1}{x^2}\right) \hat{i} \cdot \hat{i} \, \mathrm{d}x\] This simplifies to \[V = -\int_{10 \, cm}^{20 \, cm} \frac{1}{x^2}\, \mathrm{d}x\]

Step 3: Evaluate the Integral and Convert Units

To evaluate the integral, first convert the limits from cm to meters: \(x_1 = 10 \, cm = 0.1 \, m\) and \(x_2 = 20 \, cm = 0.2 \, m\) Now, evaluate the integral: \[V = -\int_{0.1}^{0.2} \frac{1}{x^2} \, \mathrm{d}x = -\left[-\frac{1}{x}\right]_{0.1}^{0.2} = (-\frac{1}{0.2}) - (-\frac{1}{0.1}) = -5 + 10 = 5 \, V\]

Step 4: Find the Magnitude of the Potential Difference

Finally, the magnitude of the potential difference is the absolute value of the result from step 3: \[|V| = |5 \, V| = 5 \, V\] Thus, the correct option is (C) \(5 \mathrm{~V}\).

Key Concepts

Electric Field

Understanding the electric field is crucial in the study of electric potential difference. The electric field, denoted by \(\vec{E}\), is a vector field that represents the force exerted per unit charge at any point in space. Think of it as a way to quantify the influence exerted by electric charges on their surroundings. In this exercise, the electric field is given by \(\vec{E}=\left(\frac{1}{x^{2}}\right) \hat{i}\, \mathrm{V} / \mathrm{m}\). This indicates that the field's strength decreases with the square of the distance \(x\). Such an inverse-square relationship is typical in force fields that radiate from a point source, similar to gravity or light.

The direction of the electric field is represented by \(\hat{i}\), the unit vector in the \(x\)-direction, meaning the field acts along the x-axis. This establishes the path along which we need to compute potential differences. By understanding that electric fields are gradients of potential, we can see why they are tied intimately to electric potential differences – they show how the potential energy landscape changes in space.

In essence, if you can understand the electric field, you can predict how charged objects will move and how potential differences arise.

Integration of Electric Field

Integration is key to finding the electric potential difference between two points. Integration, in this context, involves summing the electric field along a path. This computation gives us the potential difference through the relationship:

• \(V = -\int_{x_1}^{x_2} \vec{E} \cdot \mathrm{d}\vec{r}\)

The negative sign is essential because it indicates that potential decreases in the direction of the field.

For our specific problem, we integrate the field along the x-axis, where \(\vec{E} = \left(\frac{1}{x^2}\right) \hat{i}\), from \(x_1 = 10 \text{ cm}\) to \(x_2 = 20 \text{ cm}\). Integration requires carefully considering the limits and ensuring units are consistent. By converting the cm into meters, we can correctly apply the integral:

• First, express limits in meters: \(x_1 = 0.1\, \mathrm{m}\), \(x_2 = 0.2\, \mathrm{m}\).
• Integrate \(\int_{0.1}^{0.2} \frac{1}{x^2} \, \mathrm{d}x\).
• Evaluate to get \(-\left[-\frac{1}{x}\right]_{0.1}^{0.2} = 5\, \mathrm{V}\).

Each step ensures we properly integrate the electric field, leading us to the sought potential difference.

Integral Calculus in Physics

Integral calculus is a powerful tool in physics, often used to determine potential differences, accumulated charges, or the total work done by a force. In this exercise, integral calculus enabled us to transition from the electric field to the electric potential difference, a key concept in electrostatics. Calculus allows us to manage continuous variables and curve-like paths through integration.

This particular case focused on calculating a single-variable integral due to the dependence of \(\vec{E}\) only on the \(x\)-variable. Integral calculus helps bridge the gap between local field effects at infinitesimal points and global properties like potential difference. In future problems, you'll find integration useful when:

• Electric fields vary over space.
• Complex geometries necessitate path integrals.
• Computations require non-uniform charge distributions.

Mastering these integrations builds a more robust understanding of physics by enabling you to predict phenomena from theoretical equations and derive meaningful relationships between different physical quantities.