Problem 89 - A Complete Resource Book in Physics for JEE Main

Question

The electric potential \(V\) at any point \(x, y, z\) (all in metres) in space is given by \(V=4 x^{2}\) volts. The electric field (in \(\mathrm{V} / \mathrm{m}\) ) at the point ( \(1 \mathrm{~m}, 0,2 \mathrm{~m}\) ) (A) \(-8 \hat{i}\) (B) \(8 \hat{i}\) (C) \(-16 \hat{i}\) (D) \(8 \sqrt{5} \hat{i}\)

Step-by-Step Solution

Answer

The electric field E at point (1 m, 0, 2 m) is -8\(\hat{i}\).

Step 1: Find the partial derivatives of V

To find the electric field, we need to determine the partial derivative of V with respect to x, y, and z. Here, V is given as \(V(x, y, z)=4x^2\). The partial derivatives are: \( \frac{\partial V}{\partial x} = \frac{d (4x^{2})}{dx} = 4(2x)= 8x\) \(\frac{\partial V}{\partial y} = \frac{d(4x^{2})}{dy} = 0\) (since there is no y component in the equation) \(\frac{\partial V}{\partial z} = \frac{d(4x^{2})}{dz} = 0\) (since there is no z component in the equation)

Step 2: Find the electric field using ∇V

Now that we have the partial derivatives, we can compute the electric field E at point (1 m, 0, 2 m). We use the formula E = -∇V: E = -\(\left(\frac{\partial V}{\partial x}, \frac{\partial V}{\partial y}, \frac{\partial V}{\partial z}\right)\) = -\((8x, 0, 0)\) At point (1 m, 0, 2 m): E = -\((8\cdot1, 0, 0)\) = -\((8, 0, 0)\)

Step 3: Represent the electric field in terms of unit vectors

Now we need to represent our electric field E as a vector using unit vectors \(\hat{i}, \hat{j},\) and \(\hat{k}\): E = -8\(\hat{i}\) + 0\(\hat{j}\) + 0\(\hat{k}\) = -8\(\hat{i}\) The correct answer is (A) -8\(\hat{i}\).

Key Concepts

Electric Field

The electric field is a crucial concept in physics, especially when dealing with electric potential and forces. It represents the force experienced by a unit positive charge placed in the electric field. The relationship between electric potential, denoted as \(V\), and the electric field \(\mathbf{E}\) is significant. The electric field is basically the gradient of the electric potential. Mathematically, it is expressed as \( \mathbf{E} = - abla V \). Here, \(abla V\) signifies the gradient of the potential, which gives the direction of the maximum rate of increase of \(V\). By taking the negative sign, \(\mathbf{E}\) points in the direction of the steepest decrease in potential.
So, calculating the electric field involves determining how the potential changes in three dimensions, which is where derivatives come into play. Think of it like a hill where the potential is represented by height. The steeper the hill, the stronger the electric field; this tells us in which way the potential decreases fastest.
Understanding these basics helps us solve problems involving electric fields, such as determining the direction and magnitude of forces on charges.

Partial Derivatives

When dealing with functions of several variables, partial derivatives are key to understanding how the function changes in each direction. The electric potential \(V\) given in the exercise was \(V(x, y, z) = 4x^2\), which clearly only depends on \(x\). Hence, to find the electric field, we compute the partial derivatives with respect to \(x, y,\) and \(z\).

• For \(\frac{\partial V}{\partial x}\), we calculate how \(V\) changes with a slight change in \(x\), keeping \(y\) and \(z\) constant. Here it results in \(8x\).
• For \(\frac{\partial V}{\partial y}\), \(V\) doesn't change with \(y\) since there is no \(y\) in the equation, giving us 0.
• Similarly, \(\frac{\partial V}{\partial z}\) is also 0.

Partial derivatives help us break down complex changes in a multi-dimensional space, enabling us to understand the behavior of each dimension separately. After this, we piece these derivatives together to understand the multi-variable function's behavior in our space.

Vector Representation

Vectors are essential in physics, as they offer a clear way to represent quantities that have both magnitude and direction, such as the electric field. In this exercise, once the components of the electric field were calculated from partial derivatives, it was necessary to express the electric field vector in terms of the standard unit vectors \(\hat{i}, \hat{j},\) and \(\hat{k}\).
For the given problem, the electric field \(\mathbf{E}\) was determined as \(-8\) in the \(x\)-direction while being zero in the \(y\) and \(z\) directions. Vector representation consolidates these into a single, elegant expression: \(-8\hat{i} + 0\hat{j} + 0\hat{k} = -8\hat{i}\).

• \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\) are unit vectors along the \(x\), \(y\), and \(z\) axes, respectively.
• This method not only shows the direction but also scales the field appropriately, making it easier to visualize and work with in calculations.

This concise representation is powerful as it highlights the direction and magnitude of the electric field in a clear, manageable form.