Problem 152 - Master Resource Book in JEE Main Physics

Question

The electric potential \(V\) at any point \(O(x, y, z\) all in metres) in space is given by \(V=4 x^{2}\) volt. The electric field at the point \((1 \mathrm{~m}, 0,2 \mathrm{~m})\) in \(\mathrm{V} / \mathrm{m}\) is (a) 8 along negative \(x\)-axis (b) 8 along positive \(x\)-axis (c) 16 along negative \(x\)-axis (d) 16 along positive \(x\)-axis

Step-by-Step Solution

Answer

The electric field is 8 V/m along the negative x-axis (option a).

Step 1: Understand the relationship between electric potential and electric field

The electric field, \( \mathbf{E} \), is the negative gradient of the electric potential, \( V \). Mathematically, this is represented as \( \mathbf{E} = -abla V \). The components of \( \mathbf{E} \) can be found by calculating the partial derivatives of \( V \) with respect to \( x, y, \) and \( z \).

Step 2: Compute the partial derivatives of V

Given the potential \( V = 4x^2 \), we calculate its gradient components. Since \( V \) is a function of \( x \) only:- \( \frac{\partial V}{\partial x} = 8x \)- \( \frac{\partial V}{\partial y} = 0 \)- \( \frac{\partial V}{\partial z} = 0 \)Thus, \( abla V = (8x, 0, 0) \).

Step 3: Evaluate the electric field at the given point

The point given is \((1 m, 0, 2 m)\). Plug \( x = 1 \) into the gradient expression: \( abla V = (8 \times 1, 0, 0) = (8, 0, 0) \).

Step 4: Determine the direction and magnitude of the electric field

The electric field \( \mathbf{E} = -abla V = (-8, 0, 0) \) has a magnitude of 8, directed along the negative \( x \)-axis since the component is negative.

Key Concepts

Electric Potential

Electric potential is a fundamental concept in electromagnetism that describes the potential energy a charged particle experiences in an electric field. You can think of electric potential as a way to understand how much work is needed to move a charge within the field. It is often denoted by the symbol \( V \) and is measured in volts (V).

This potential arises due to the presence of electric charges and indicates how the charges can influence other charged objects nearby. The potential itself is a scalar quantity. This means it has a magnitude but no direction.

In the context of our exercise, the electric potential is given by the equation \( V = 4x^2 \). This indicates the potential varies with the position \( x \) in space. It highlights how potential energy changes for a set of points in a defined space, particularly when the potential is defined only as a function of one variable like \( x \) in this exercise.

Gradient of Potential

The gradient of potential is a crucial concept for linking potential fields with the forces experienced by charges. In simple terms, the gradient refers to the rate and direction of the steepest increase of the function, which in this case is electric potential \( V \).

The negative gradient of the electric potential gives us the electric field \( \mathbf{E} \). This is mathematically expressed as \( \mathbf{E} = -abla V \). Here, \( abla V \) is a vector that shows how much and in what direction the potential changes at each point in space.

When computing the gradient for our potential \( V = 4x^2 \), the results show that changes in potential only occur along the \( x \)-axis. Thus, \( abla V = (8x, 0, 0) \), indicating a change only in the \( x \) direction. The direction of the electric field is then the inverse direction of this gradient, aligning with how fields direct forces on charges.

Partial Derivatives

Partial derivatives are a powerful tool used to calculate the change of a function with respect to one variable while keeping the others constant. This is important when dealing with functions of multiple variables, as is often the case in electromagnetism.

For the electric potential given, \( V = 4x^2 \), we need to compute its partial derivatives with respect to \( x, y, \) and \( z \).

• \( \frac{\partial V}{\partial x} = 8x \) — This helps determine how the potential changes as we vary \( x \), showing a growth rate proportionate to \( x \).
• \( \frac{\partial V}{\partial y} = 0 \) — Indicates no change in potential when changing \( y \).
• \( \frac{\partial V}{\partial z} = 0 \) — Indicates no change in potential when changing \( z \).

These results show the potential is only dependent on \( x \). The gradient thus forms a vector \( (8x, 0, 0) \), revealing the structure of the potential field and guiding how we compute the electric field.