Problem 78 - University Physics with Modern Physics

Question

The electric potential V in a region of space is given by $$V(x, y, z) = A(x^2 - 3y^2 + z^2)$$ where \(A\) is a constant. (a) Derive an expression for the electric field \(\overrightarrow{E}\) at any point in this region. (b) The work done by the field when a 1.50-\(\mu\)C test charge moves from the point \((x, y, z) = (0, 0, 0.250 m)\) to the origin is measured to be 6.00 \(\times 10^{-5}\) J. Determine A. (c) Determine the electric field at the point \((0, 0, 0.250 m)\). (d) Show that in every plane parallel to the \(xz\)-plane the equipotential contours are circles. (e) What is the radius of the equipotential contour corresponding to \(V = 1280\) \(V\) and \(y = 2.00\) m?

Step-by-Step Solution

Answer

(a) \(\overrightarrow{E} = -2Ax \hat{i} + 6Ay \hat{j} - 2Az \hat{k}\). (b) A = -640 V/m². (c) \(\overrightarrow{E} = 320 \hat{k}\) V/m. (d) Contours are circles. (e) Radius = 2 m.

Step 1: Calculate Electric Field from Potential

The electric field \( \overrightarrow{E} \) is related to the electric potential \( V \) by the negative gradient: \( \overrightarrow{E} = -abla V \). Calculate each component of the gradient: \[ abla V = \left( \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y}, \frac{\partial V}{\partial z} \right) \].First, compute \( \frac{\partial V}{\partial x} = 2Ax \). Next, compute \( \frac{\partial V}{\partial y} = -6Ay \). Finally, compute \( \frac{\partial V}{\partial z} = 2Az \). Thus, the electric field is: \[ \overrightarrow{E} = -\left( 2Ax \hat{i} - 6Ay \hat{j} + 2Az \hat{k} \right) = -2Ax \hat{i} + 6Ay \hat{j} - 2Az \hat{k} \].

Step 2: Determine Constant A

Use the work-energy principle, which states that the work done \( W \) by the electric field is equal to \(- \Delta V \cdot q \), where \( q \) is the charge. Given that \( W = 6.00 \times 10^{-5} \) J and \( q = 1.50 \times 10^{-6} \) C, set \( \Delta V = V(0, 0, 0.250) - V(0, 0, 0) = A((0)^2 - 3(0)^2 + 0.250^2) - 0 = A(0.0625) \).Then, \(-A(0.0625) \times 1.50 \times 10^{-6} = 6.00 \times 10^{-5} \), solve for \( A \): \[ A = -\frac{6.00 \times 10^{-5}}{0.0625 \times 1.50 \times 10^{-6}} = -640 \] V/m².

Step 3: Calculate Electric Field at Specific Point

Using the expression for \( \overrightarrow{E} \) found earlier: \[ \overrightarrow{E} = -2Ax \hat{i} + 6Ay \hat{j} - 2Az \hat{k} \]. Substitute \( x = 0, y = 0, z = 0.250 \) m, and \( A = -640 \) into the equation: \[ \overrightarrow{E}(0, 0, 0.250) = -2(0)(-640) \hat{i} + 6(0)(-640) \hat{j} - 2(0.250)(-640) \hat{k} = 320 \hat{k} \].Thus, \( \overrightarrow{E} = 320 \hat{k} \) V/m.

Step 4: Equipotential Contours in the xz-plane

In a plane parallel to the \( xz \)-plane (i.e., \( y \) is constant), the potential \( V(x, 0, z) = A(x^2 + z^2) \) resembles the equation of a circle \( x^2 + z^2 = \frac{V}{A} \), indicating circular equipotential lines. Thus, in every plane parallel to the \( xz \)-plane, equipotential contours form circles centered on the origin.

Step 5: Determine Radius of Equipotential Circle

For a constant \( V \) and given \( y = 2.00 \) m, use the equation \( V = A(x^2 - 3 \times 4 + z^2) = x^2 + z^2 - 12A \). Solve for radius using \( x^2 + z^2 = \frac{1280 + 12(-640)}{-640} = 4 \). Thus, the radius \( r \) is \( \sqrt{4} = 2 \) m.

Key Concepts

Electric Field

An electric field describes the force experienced by a charge in space due to the presence of other charges. It is a vector field, meaning it has both magnitude and direction.
The relation between the electric field \( \overrightarrow{E} \) and the electric potential \( V \) is given by the negative gradient of the potential: \( \overrightarrow{E} = -abla V \). This means the electric field points in the direction where the potential decreases most rapidly and calculates the rate of this change.
In the problem, we calculate each component of the gradient of potential \( V(x, y, z) = A(x^2 - 3y^2 + z^2) \):

• The partial derivative with respect to \( x \): \( \frac{\partial V}{\partial x} = 2Ax \)
• The partial derivative with respect to \( y \): \( \frac{\partial V}{\partial y} = -6Ay \)
• The partial derivative with respect to \( z \): \( \frac{\partial V}{\partial z} = 2Az \)

By combining these, we get \( \overrightarrow{E} = -2Ax \hat{i} + 6Ay \hat{j} - 2Az \hat{k} \). This vector shows how the electric field strength and direction change with position in space.

Gradient of Potential

The gradient of potential is a spatial derivative that shows the directional rate of change of potential in a field. It is a crucial concept that links the electric potential to the electric field.
In mathematical terms, the gradient of a potential \( V \) is expressed as \( abla V = \left( \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y}, \frac{\partial V}{\partial z} \right) \). This provides the rate and direction at which the potential changes.

For the provided function \( V(x, y, z) \), we derive the components of the electric field:\

• \( \frac{\partial V}{\partial x} = 2Ax \)
• \( \frac{\partial V}{\partial y} = -6Ay \)
• \( \frac{\partial V}{\partial z} = 2Az \)

Plugging in the derivatives into our gradient, the resultant electric field is \( \overrightarrow{E} = -2Ax \hat{i} + 6Ay \hat{j} - 2Az \hat{k} \).
This expression shows how the potential directly influences the electric field, indicating a larger potential change results in a stronger electric field.

Equipotential Contours

Equipotential contours are imaginary lines or surfaces in space where the potential is constant. These contours help to visualize the electric field, as no work is done when moving a charge along these lines.
For the function \( V(x, y, z) = A(x^2 - 3y^2 + z^2) \), consider when \( y = 0 \); this simplifies to \( V = A(x^2 + z^2) \), which represents a circle in the \( xz \)-plane.
This means the potential value remains the same for all points on the circle. Such circular paths are an indication of uniform field lines in regions with symmetry.
The radius of these equipotential contours can be determined using the formula \( x^2 + z^2 = \frac{V}{A} \), emphasizing how contours adjust with different potential values.

Work-Energy Principle

The work-energy principle connects the work done by a field to changes in potential energy and is key to calculating the constant \( A \) in this problem.
The formula used is \(-\Delta V \cdot q = W\), where \( W \) is work, \( \Delta V \) is the potential difference, and \( q \) is the charge involved. This principle tells us that the work done by the field when a charge moves between two points equates to the change in potential energy.
From the exercise, as the 1.50-\( \mu \)C charge moves from one point to another with a given work done, these values help find the potential difference. Calculating \( A \) involves considering \( A(0.0625) = -\frac{W}{q} \), indicating that the work done is directly linked to potential change, ultimately determining the required constant \( A = -640 \) V/m².
Therefore, understanding this principle aids in the accurate analysis of electric fields and potentials.