Problem 78 - Master Resource Book in JEE Main Physics

Question

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

Step-by-Step Solution

Answer

(a) 8, along negative X-axis.

Step 1: Understanding the Relationship Between Electric Potential and Electric Field

The electric field \( \mathbf{E} \) is related to the electric potential \( V \) by the equation \( \mathbf{E} = -abla V \). This means the electric field is the negative gradient of the electric potential.

Step 2: Calculate the Gradient of the Electric Potential

Given the electric potential \( V = 4x^2 \), we need to find \( abla V \). For the gradient, compute the partial derivatives: abla V = \left( \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y}, \frac{\partial V}{\partial z} \right).

Step 3: Compute Partial Derivatives

Calculate each partial derivative: \( \frac{\partial V}{\partial x} = 8x \), \( \frac{\partial V}{\partial y} = 0 \), and \( \frac{\partial V}{\partial z} = 0 \). Thus, \( abla V = (8x, 0, 0) \).

Step 4: Evaluate the Gradient at the Given Point

Substitute the point \((1,0,2)\) into the gradient: \( abla V \) at \((1,0,2)\) is \((8 \times 1, 0, 0) = (8, 0, 0)\).

Step 5: Calculate the Electric Field

Using \( \mathbf{E} = -abla V \), substitute the gradient at \((1,0,2)\): \( \mathbf{E} = -(8, 0, 0) = (-8, 0, 0)\).

Step 6: Determine the Direction and Magnitude of the Electric Field

The electric field \( (-8, 0, 0) \) has a magnitude of \( 8 \) and is in the direction of the negative X-axis.

Key Concepts

Electric Potential

Electric potential, often represented by the symbol \( V \), is a fundamental concept in electromagnetism. It describes the potential energy per unit charge at any point in an electric field. Think of it as the capacity for doing work due to the position of a charge relative to other charges.
- **Energy Perspective**: The electric potential indicates how much work is needed to move a charge into a specific position relative to an electric field.- **Expression in Formulas**: It is usually given in volts (V) and can be determined by integrating the electric field. In this exercise, we are given the electric potential as \( V = 4x^2 \), a function dependent on the \( x \)-coordinate. Understanding electric potential is crucial because it helps in calculating the electric field, a vector field representing forces acting on charges.

Gradient

The gradient is a vector field that shows how a scalar field, like electric potential, changes in space. It points in the direction of the greatest rate of increase of the function.
- **Calculating Gradients**: To find the gradient of a potential, compute the partial derivatives of the potential with respect to each variable. For the function \( V = 4x^2 \): \( \frac{\partial V}{\partial x} = 8x, \quad \frac{\partial V}{\partial y} = 0, \quad \frac{\partial V}{\partial z} = 0 \) - **Resulting Vector**: This gives us the vector \( abla V = (8x, 0, 0) \), which denotes how \( V \) changes in each coordinate direction. The gradient tells us how the potential influences movement of charges in the field. It's vital for calculating how the potential energy translates into actual motion or force.

Partial Derivative

Partial derivatives are used in multivariable calculus to determine the rate at which a function changes along one variable, with others held constant. This is crucial when examining fields like the electric field where potential is defined in three dimensions.
- **Purpose**: By taking partial derivatives of the electric potential \( V = 4x^2 \), we can find how \( V \) changes as we vary \( x \), while keeping \( y \) and \( z \) constant.- **Example Calculations**: - \( \frac{\partial V}{\partial x} = 8x \) (captures how \( V \) changes with \( x \)) - \( \frac{\partial V}{\partial y} = 0 \) and \( \frac{\partial V}{\partial z} = 0 \) show that \( V \) doesn't change with \( y \) or \( z \). These derivatives are essential for gradient calculations, ultimately leading to the determination of the electric field, which guides the movement of charges.