Question

The electric potential \(\mathrm{V}\) at any point \(\mathrm{x}, \mathrm{y}, \mathrm{z}\) (all in meter) 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 \(\mathrm{Vm}^{-1}\) is \((\mathrm{A})+8 \mathrm{i} \wedge\) (B) \(-8 \mathrm{i} \wedge\) (C) \(-16 \mathrm{i}\) (D) \(+16 \mathrm{i}\)

Short answer

The electric field at point \((1\ \mathrm{m}, 0, 2\ \mathrm{m})\) is \((B)\ -8\vec{i}\ \mathrm{Vm}^{-1}\).

Step-by-step solution

Find the derivative of the potential function with respect to x

Given \(V = 4x^2\) (y and z do not affect V). We will differentiate V with respect to x: \[ \frac{\partial V}{\partial x} = \frac{d(4x^2)}{dx} = 8x \]

Calculate the x-component of the electric field Ex using the formula E = -∇V

The x-component of the electric field Ex is given by: \[ Ex = -\frac{\partial V}{\partial x} = -8x \]

Evaluate Ex at the given point (1m, 0, 2m)

To find the electric field at the point (1m, 0, 2m), we plug in x=1 into Ex: \[ Ex = -8(1) = -8\ \mathrm{Vm}^{-1} \] So, the electric field at the given point is -8 Vm^{-1} in the x-direction. Checking the given options, we can conclude that the correct answer is: (B) \(-8\vec{i}\ \mathrm{Vm}^{-1}\)

Key concepts

Electric Potential

Electric potential is a way to quantify the amount of electric potential energy per unit charge at a point in space due to a static electric field. It is often denoted by the symbol \( V \) and is measured in volts (\( V \)).
In simple terms, it helps us understand how much work would be needed to move a charged particle to a certain point in an electric field without speed change. In this exercise, the electric potential, \( V \), is given by the equation \( V = 4x^2 \). This means the potential varies with the square of \( x \) while \( y \) and \( z \) do not contribute to it.
When working with electric potential, it’s crucial to understand that the change in potential energy leads directly to an electric field, a vector quantity that shows the force exerted on charges in the field.

Vector Calculus

Vector calculus is a branch of mathematics concerned with differentiation and integration of vector fields. In physics, it’s integral to understanding electric fields and potentials, which are often vector valued functions.
When dealing with electric potential, the gradient operator \( abla \) is a key element in vector calculus. It helps find the rate and direction of change. In mathematics, the negative gradient of a potential function gives you the corresponding electric field vector.

• The gradient \( abla V \) is calculated by taking partial derivatives concerning each spatial coordinate.
• The negative sign before \( abla V \) ensures the electric field points in the direction of decreasing potential.

In this context, understanding vector calculus allows us to link the scalar potential function with the vector electric field.

Partial Derivative

A partial derivative is a derivative where one variable is differentiated while all other variables are held constant. This is particularly important in physics for understanding how a function changes in relation to just one of its inputs.
From the exercise, we see the potential function depends on \( x \) but not \( y \) or \( z \), so the partial derivative \( \frac{\partial V}{\partial x} \) gives us how \( V \) changes with small changes in \( x \).

• Given \( V = 4x^2 \), the calculation \( \frac{d(4x^2)}{dx} = 8x \) demonstrates this for \( x \).
• This partial derivative is used in calculations to determine the x-component of the electric field in the problem.

Partial derivatives are crucial when variables are interdependent but affect functions differently.

Physics Problem Solving

Physics problem solving involves taking physical scenarios and breaking them into understandable mathematical problems. Often in electromagnetic problems, such as this one, the solution requires using mathematical tools to calculate quantities like electric fields from potentials.

• You start by identifying what you know: the potential \( V \) in terms of \( x \).
• You then apply mathematical operations, calculating partial derivatives to account for rate of changes as needed.
• Finally, you use these calculated values to determine related properties at a specified point, such as the electric field.

In this task, after deriving the partial derivative and recognizing the connection between the negative gradient and electric fields, you arrive at a precise value for the electric field at the point \( (1m, 0, 2m) \). This systematic approach ensures accurate solutions to often complex physics tasks.