Problem 46 - University Physics with Modern Physics

Question

In a certain region of space the electric potential is given by \(V=+A x^{2} y-B x y^{2},\) where \(A=5.00 \mathrm{V} / \mathrm{m}^{3}\) and \(B=\) 8.00 \(\mathrm{V} / \mathrm{m}^{3} .\) Calculate the magnitude and direction of the electric field at the point in the region that has coordinates \(x=2.00 \mathrm{m}\) \(y=0.400 \mathrm{m},\) and \(z=0\).

Step-by-Step Solution

Answer

The magnitude of the electric field is approximately 9.85 V/m, and its direction is 47.94° from the positive x-axis towards the fourth quadrant.

Step 1: Understand the Relationship

The electric field \(\mathbf{E}\) is related to the electric potential \(V\) by the negative gradient, i.e., \( \mathbf{E} = -abla V \). At the point of interest, calculate the partial derivatives of \(V\) with respect to each coordinate to find \(E_x\), \(E_y\), and \(E_z\).

Step 2: Calculate Partial Derivatives

Compute the partial derivative with respect to \(x\): \( \frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(A x^2 y - B x y^2) = 2Axy - By^2 \).Compute the partial derivative with respect to \(y\): \( \frac{\partial V}{\partial y} = \frac{\partial}{\partial y}(A x^2 y - B x y^2) = Ax^2 - 2Bxy \).For \(z\), the potential is independent of \(z\), so \( \frac{\partial V}{\partial z} = 0 \).

Step 3: Substitute Coordinates for Electric Field Components

Substitute \(x = 2.00 \, \text{m}\) and \(y = 0.400 \, \text{m}\) into the partial derivatives to find \(E_x\) and \(E_y\).Calculate \( E_x = - (2Axy - By^2) = -(2(5.00)(2.00)(0.400) - 8.00(0.400)^2) \).Calculate \( E_y = - (Ax^2 - 2Bxy) = -(5.00(2.00)^2 - 2(8.00)(2.00)(0.400)) \).Since \( \frac{\partial V}{\partial z} = 0 \), \( E_z = 0 \).

Step 4: Evaluate the Electric Field Components

Calculate \( E_x \):\( E_x = -(2(5.00)(2.00)(0.400) - 8.00(0.400)^2) = -(8.00 - 1.28) = -6.72 \, \text{V/m} \).Calculate \( E_y \):\( E_y = -(5.00(2.00)^2 - 2(8.00)(2.00)(0.400)) = -(20.00 - 12.80) = -7.20 \, \text{V/m} \).So, \( E_z = 0 \) V/m because there's no dependency on \(z\).

Step 5: Calculate Magnitude and Direction of Electric Field

Find the magnitude of the electric field using the components calculated: \( |\mathbf{E}| = \sqrt{E_x^2 + E_y^2} = \sqrt{(-6.72)^2 + (-7.20)^2} \).Evaluate the magnitude: \( |\mathbf{E}| = \sqrt{45.1584 + 51.84} = \sqrt{96.9984} \approx 9.85 \, \text{V/m} \).Determine the direction by calculating the angle \((\theta)\): \( \theta = \text{atan}\left(\frac{E_y}{E_x}\right) = \text{atan}\left(\frac{-7.20}{-6.72}\right) \).Calculate the angle: \( \theta \approx 47.94^\circ \) from the positive \(x\) axis in the fourth quadrant, because both components are negative.

Key Concepts

Electric Potential

Imagine a gentle, invisible push around an area in space. This push is what we refer to as electric potential. It is essentially a measure of the potential energy per unit charge at any point in a field. In simpler terms, think of it like elevation maps showing hills and valleys; only here, electric potential depicts regions of high and low electric energy.

In our exercise, we have an equation giving the electric potential, \( V = Ax^2y - Bxy^2 \). Here, \(A\) and \(B\) are constants that determine the scale of the potential. The equation specifies how the potential changes over space with respect to the coordinates \(x\) and \(y\).

• The potential is shaped by the values of \(A = 5.00 \, \mathrm{V/m}^3\) and \(B = 8.00 \, \mathrm{V/m}^3\).
• This equation helps us determine behavior of charges, guiding them through the electric field, much like a river follows a valley.

Gradient of a Scalar Field

The concept of a gradient is crucial when dealing with fields described by a scalar quantity, like electric potential. Essentially, the gradient is a vector that points in the direction of the steepest increase of a function, and whose magnitude represents how much the function rises in that direction.

In this exercise, we derive the electric field, \( \mathbf{E} \), from the electric potential, using the negative gradient of \(V\). This is written mathematically as \( \mathbf{E} = -abla V \). The gradient operator \(abla\) is a fancy way of saying "find how the function changes for each coordinate". This operator brings together partial derivatives to form a vector, showing how quickly and in which direction the potential changes.

• This method helps us find the electric field, which affects how charged particles move.
• The gradient captures both the strength and direction of the field, unlike scalar potential which only measures strength.

Partial Derivatives

Partial derivatives are essential tools in multivariable calculus. They give us insight into how a function changes with respect to one variable alone, while keeping other variables constant.

In our example, to compute the electric field, we need to determine these derivatives for the potential function \( V \). Calculating \( \frac{\partial V}{\partial x} \) and \(\frac{\partial V}{\partial y}\), involves focusing on how \(V\) alters when each coordinate changes independently.

Here's how the process works:

• Take \( \frac{\partial V}{\partial x} = 2Axy - By^2 \) as the rate of change of potential with respect to \(x\). This gives us insight into the \(x\)-component of the electric field, \(E_x\).
• Next, \(\frac{\partial V}{\partial y} = Ax^2 - 2Bxy\) shows how the potential changes along the \(y\)-direction. This helps to find \(E_y\).

These derivatives are deterministic in finding the force experienced by a charge due to changes in potential at any given point.

Electric Field Components

The electric field itself is often described in terms of components, which break down the field into its effects in the directions of the coordinate axes. Think of it like dissecting a force into its vertical and horizontal pushes.

For our problem, we consider the components \(E_x\), \(E_y\), and \(E_z\):

• \(E_x\) and \(E_y\) are calculated using the negative of the partial derivatives of \(V\) concerning \(x\) and \(y\), respectively.
• Since the potential \(V\) does not change with \(z\), \(E_z = 0\).
  
  When we substitute the given point's coordinates \((x = 2.00 \, \mathrm{m}, y = 0.400 \, \mathrm{m})\) into \(E_x = -(2Axy - By^2)\) and \(E_y = -(Ax^2 - 2Bxy)\), we obtain specific values for \(E_x\) and \(E_y\), which tell us how strong the field is in each direction.

These components then allow us to calculate the field's magnitude and direction, fully describing the push the field exerts.