Question

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

Short answer

The electric field magnitude is approximately 9.85 V/m, directed at about 47.43° from the negative x-axis towards the negative y-axis.

Step-by-step solution

Understanding the electric potential function

The electric potential is given by the function \( V = +Ax^2y - Bxy^2 \), where \( A = 5.00 \, V/m^3 \) and \( B = 8.00 \, V/m^3 \). We need to find the electric field, which is related to the potential by the negative gradient \( \textbf{E} = -abla V \).

Calculating the partial derivatives of the potential

The electric field components \( E_x \), \( E_y \), and \( E_z \) are given by:\[ E_x = -\frac{\partial V}{\partial x} \]\[ E_y = -\frac{\partial V}{\partial y} \]\[ E_z = -\frac{\partial V}{\partial z} \]Since \( V \) is not dependent on \( z \), \( E_z = 0 \).

Compute the x-component of the electric field

Find \( \frac{\partial V}{\partial x} \):\[ \frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(Ax^2y - Bxy^2) = 2Axy - By^2 \]Substitute \( A = 5.00 \, V/m^3 \), \( B = 8.00 \, V/m^3 \), \( x = 2.00 \, m \), and \( y = 0.400 \, m \):\[ \frac{\partial V}{\partial x} = 2(5.00)(2.00)(0.400) - 8.00(0.400)^2 = 8.00 - 1.28 = 6.72 \, V/m \]Thus, \( E_x = -6.72 \, V/m \).

Compute the y-component of the electric field

Find \( \frac{\partial V}{\partial y} \):\[ \frac{\partial V}{\partial y} = \frac{\partial}{\partial y}(Ax^2y - Bxy^2) = Ax^2 - 2Bxy \]Substitute \( A = 5.00 \, V/m^3 \), \( B = 8.00 \, V/m^3 \), \( x = 2.00 \, m \), and \( y = 0.400 \, m \):\[ \frac{\partial V}{\partial y} = (5.00)(2.00)^2 - 2(8.00)(2.00)(0.400) = 20.00 - 12.80 = 7.20 \, V/m \]Thus, \( E_y = -7.20 \, V/m \).

Calculating the magnitude and direction of the electric field

The magnitude of the electric field \( E \) is given by:\[ |\textbf{E}| = \sqrt{E_x^2 + E_y^2 + E_z^2} = \sqrt{(-6.72)^2 + (-7.20)^2 + 0^2} = \sqrt{45.1584 + 51.84} = \sqrt{96.9984} \approx 9.85 \, V/m \]The direction can be found by:\( \theta = \tan^{-1}\left( \frac{E_y}{E_x} \right) = \tan^{-1}\left( \frac{-7.20}{-6.72} \right) \approx \tan^{-1}(1.071) \approx 47.43^\circ \) from the negative x-axis towards the negative y-axis.

Key concepts

Electric Potential

Electric potential is a measure of the work needed to move a unit charge from a reference point to a specific point within an electric field without any acceleration. It is a scalar quantity that represents the potential energy per unit charge. In our exercise, the electric potential is given by the equation:
\( V = +Ax^2y - Bxy^2 \), where \( A = 5.00 \, V/m^3 \) and \( B = 8.00 \, V/m^3 \).
Electric potential can change based on the position within the field

• As shown here, the potential is dependent on the coordinates \( x \) and \( y \), and the constants \( A \) and \( B \), which are specific characteristics of this potential field.
• An understanding of how potential varies is critical because the electric field is derived from the spatial change of this potential.

The concept of electric potential simplifies the mathematics of electromagnetism by reducing vector calculations to simpler scalar operations.

Partial Derivatives

Partial derivatives are used to measure how a function changes as one of the variables changes while holding the other variables constant.
In the context of electric potential, partial derivatives allow us to calculate the components of the electric field from the scalar potential.
For this problem, we need the partial derivatives of \( V \) with respect to \( x \) and \( y \), which correspond to the components \( E_x \) and \( E_y \), respectively:

• \( \frac{\partial V}{\partial x} = 2Axy - By^2 \)
• \( \frac{\partial V}{\partial y} = Ax^2 - 2Bxy \)

These derivatives help us understand how swiftly the potential changes in space, which is essential for determining the electric field direction and strength in any specific spot.

Gradient

The gradient is a vector operation that takes a scalar field, like electric potential, and returns a vector field, which in this case is the electric field. This process is critical because it provides information about both the magnitude and the direction of the field.
The electric field \( \bf{E} \) is related to the potential \( V \) through the negative gradient:

• \( \bf{E} = -abla V \)

The gradient essentially points in the direction of the greatest rate of increase of the scalar field, but we take the negative for electric fields because the field lines point from high to low potential. This vector \( \bf{E} \) tells us how the electric potential is changing spatially, providing the electric field's direction and magnitude at any point in the region.

Vector Magnitude

Vector magnitude is an essential aspect of understanding electric fields as it gives us the intensity of the field. In our problem, once we have the components of the electric field, \( E_x \), \( E_y \), and \( E_z \) (where \( E_z = 0 \)), we calculate the electric field's total magnitude using:

• \( |\bf{E}| = \sqrt{E_x^2 + E_y^2 + E_z^2} \)

The magnitude tells us how strong the electric field is at the given point.
In the given exercise, we determined:
\( E_x = -6.72 \, V/m \) and \( E_y = -7.20 \, V/m \), leading to a total magnitude:\( \approx 9.85 \, V/m \).
Understanding vector magnitude helps in visualizing just how intense the electric field is at a specific point, influencing how much force a charged particle would experience at that location.