Problem 47 - University Physics with Modern Physics

Question

In a certain region of space, the electric potential is \(V(x, y, z)=A x y-B x^{2}+C y,\) where \(A, B,\) and \(C\) are positive constants. (a) Calculate the \(x-y\) - and \(z\) -components of the electric field. (b) At which points is the electric field equal to zero?

Step-by-Step Solution

Answer

The electric field is zero at \(x = -\frac{C}{A}\) and \(y = \frac{2BC}{A^2}\).

Step 1: Understanding the Electric Field Equation

The electric field \(E\) is related to the electric potential \(V\) through the equation \(\mathbf{E} = -abla V\). This means that the components of the electric field are the negative partial derivatives of the electric potential with respect to \(x\), \(y\), and \(z\).

Step 2: Calculating the x-component

Calculate the \(x\)-component of the electric field using \(E_x = -\frac{\partial V}{\partial x}\). We have \(V(x, y, z) = Axy - Bx^2 + Cy\), so \(\frac{\partial V}{\partial x} = Ay - 2Bx\). Therefore, \(E_x = -(Ay - 2Bx) = 2Bx - Ay\).

Step 3: Calculating the y-component

Calculate the \(y\)-component of the electric field using \(E_y = -\frac{\partial V}{\partial y}\). We find \(\frac{\partial V}{\partial y} = Ax + C\). Therefore, \(E_y = -(Ax + C) = -Ax - C\).

Step 4: Calculating the z-component

Calculate the \(z\)-component of the electric field using \(E_z = -\frac{\partial V}{\partial z}\). Since \(V\) has no \(z\) dependence, \(\frac{\partial V}{\partial z} = 0\), hence \(E_z = 0\).

Step 5: Finding Points Where the Electric Field is Zero

To find where the electric field components are zero, set \(E_x = 0\) and \(E_y = 0\). First, solve \(2Bx - Ay = 0\) and \(-Ax - C = 0\). From \(-Ax - C = 0\), we find \(x = -\frac{C}{A}\). Substitute this into \(2Bx - Ay = 0\) to solve for \(y\), resulting in \(y = \frac{2BC}{A^2}\). Thus, the field is zero at \(x = -\frac{C}{A}\) and \(y = \frac{2BC}{A^2}\).

Key Concepts

Electric Potential

Electric potential is a fundamental concept in electrostatics. It represents the amount of work needed to move a unit positive charge from a reference point to a specific point inside an electric field without any change in kinetic energy.
In the context of the given exercise, the electric potential function is expressed as \(V(x, y, z) = Axy - Bx^2 + Cy\), where \(A\), \(B\), and \(C\) are constants. This equation describes how the electric potential changes with position in space.

• The electric potential is scalar, meaning it only has magnitude and no direction.
• Each point in space has a distinct electric potential value, depending on its coordinates \(x\), \(y\), and \(z\).

Electric potential helps us understand electric fields more intuitively by indicating the potential energy landscape that a charge would experience as it moves through an area.

Partial Derivatives

Partial derivatives are a mathematical tool used to explore how a function changes when one of its input variables changes, while others are held constant.
In the exercise, partial derivatives are crucial for finding the electric field components from the electric potential. The negative gradient of the potential \( \mathbf{E} = -abla V \) gives us the electric field. This means:

• \(E_x = -\frac{\partial V}{\partial x}\) gives the \(x\)-component of the electric field, derived from how potential changes with \(x\).
• \(E_y = -\frac{\partial V}{\partial y}\) gives the \(y\)-component, reflecting potential changes with \(y\).
• \(E_z = -\frac{\partial V}{\partial z}\) gives the \(z\)-component, but in this case, the potential doesn't depend on \(z\), resulting in \(E_z=0\).

The use of partial derivatives lets us capture the direction and magnitude of the electric field at any point, based on its spatial potential variation.

Zero Electric Field Points

Zero electric field points are specific locations in a space where the resultant electric field is null, meaning that a charge placed there would experience no net force.
To find these points, we need to set the electric field components to zero, as the exercise does:

• For the \(x\)-component, solve \( 2Bx - Ay = 0 \).
• For the \(y\)-component, solve \( -Ax - C = 0 \).

By solving these equations together, we determine the coordinates \(x = -\frac{C}{A}\) and \(y = \frac{2BC}{A^2}\).
These solutions indicate the specific positions where the forces from the different field components balance out to zero, resulting in no net movement for a charged particle placed at these points.
Such locations are crucial for understanding stability and equilibrium conditions in fields, as well as designing systems where charges can remain in static positions.