Question

The electric Potential \(\mathrm{V}\) at any Point \(0(\mathrm{x}, \mathrm{y}, \mathrm{z}\) all in meters \()\) in space is given by \(\mathrm{V}=4 \mathrm{x}^{2}\) volt. The electric field at the point \((1 \mathrm{~m}, 0.2 \mathrm{~m})\) in volt meter is \(\ldots \ldots .\) (A) 8 , along negative \(\mathrm{x}\) - axis (B) 8 , along positives \(\mathrm{x}\) - axis (C) 16 , along negative \(\mathrm{x}\) -axis (D) 16 , along positives \(\mathrm{x}\) -axis

Short answer

The electric field at the point (1m, 0.2m) is \(-8 \mathrm{V/m}\), along the negative x-axis. The correct answer is (A).

Step-by-step solution

Find the gradient of the electric potential V

First, we will find the gradient of the given electric potential V(x, y, z) = 4x². The gradient is a vector that contains the partial derivatives of the function with respect to each coordinate: $$ \nabla V = \left(\frac{\partial V}{\partial x},\frac{\partial V}{\partial y},\frac{\partial V}{\partial z}\right) $$ Compute the partial derivatives: $$ \frac{\partial V}{\partial x} = 8x, \quad \frac{\partial V}{\partial y} = 0, \quad \frac{\partial V}{\partial z} = 0 $$ So, the gradient of the electric potential is: $$ \nabla V = (8x, 0, 0) $$

Evaluate the gradient at the given point

Now, we will evaluate the gradient of the electric potential at the given point (1m, 0.2m): $$ \nabla V(1, 0.2) = (8(1), 0, 0) = (8, 0, 0) $$

Find the electric field

The electric field E is the negative gradient of the electric potential: $$ \textbf{E} = -\nabla V $$ So, at the point (1m, 0.2m), the electric field is: $$ \textbf{E}(1, 0.2) = -(8, 0, 0) = (-8, 0, 0) \; \text{V/m} $$

Match the answer with the given options

The electric field at the point (1m, 0.2m) is (-8, 0, 0) V/m, which means it has a magnitude of 8 V/m and points along the negative x-axis. This matches option (A) in the exercise. Therefore, the correct answer is: (A) 8, along negative x-axis

Key concepts

Electric Field

An electric field is a region around a charged particle or object where an electric force is exerted on other charged particles. The concept of an electric field helps us understand how charges interact in space without direct contact. Usually, we represent this field as a vector field, which means it has both a magnitude and a direction at every point in space.
The formula to find the electric field **E** is to take the negative of the gradient of the electric potential **V**. Mathematically, it's expressed as:

• \( \textbf{E} = -abla V \)

That minus sign denotes that the electric field points in the direction where the electric potential decreases the most. Think of it as moving downhill on a slope; energy always moves from higher to lower potential.
The electric field is crucial in determining the forces on charges and is measured in volts per meter \((\text{V/m})\). Using this understanding helps predict how charged particles will move in response to electric forces in their environment.

Gradient of a Scalar Field

The gradient of a scalar field is fundamental in vector calculus and physics. It's a vector operation that provides the direction and rate of the fastest increase of the scalar field. Think of the gradient as a tool that maps how steeply a hill climbs in each direction.
Mathematically, the gradient \( abla V \) of the electric potential \( V \) is defined by taking partial derivatives with respect to each coordinate (x, y, z), resulting in a vector:

• \( abla V = \left( \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y}, \frac{\partial V}{\partial z} \right) \)

The partial derivatives reveal how the function changes around a small point, helping us know how steeply the potential varies in different directions.
In the given problem, the electric potential is \( V = 4x^2 \) which simplifies the gradient computation because the function only varies with **x**. Thus, the gradient has the form \( (8x, 0, 0) \), indicating changes only along the x-direction.

Vector Calculus

Vector calculus is the branch of mathematics that deals with vector fields and operations on vectors. It extends calculus to two or more dimensions, allowing us to handle fields like the electric field that have direction and magnitude.
Key operations in vector calculus include:

• **Gradient**: Measures the rate and direction of change in a scalar field.
• **Divergence**: Determines the magnitude of a source or sink at a point in a vector field.
• **Curl**: Captures the rotation of a vector field around a point.

These operations are essential when working with physical phenomena that vary in space.
In our exercise, understanding how to calculate gradients and apply them to find electric fields is part of using vector calculus to solve real-world physics problems. Vector calculus allows scientists and engineers to model and predict behaviors involving fields like electromagnetism, fluid flow, and more.