Problem 86 - University Physics with Modern Physics

Question

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

Step-by-Step Solution

Answer

The electric field is \(\vec{E} = (-2Ax, 6Ay, -2Az)\). For the given movement, \(A = -640\). At \((0,0,0.250)\) m, \(\vec{E} = 320 \hat{k}\). Equipotential contours are circles, radius \(\sqrt{10}\) m at \(V=1280\) V, \(y=2\) m.

Step 1: Derive the Electric Field Equation

The electric field \( \vec{E} \) is related to the electric potential \( V \) by the negative gradient, i.e., \( \vec{E} = - abla V \). The gradient in Cartesian coordinates is given by \( abla = \left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right) \). Therefore,\[ \vec{E} = - \left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} \right) \]Substituting \( V(x, y, z) = A(x^2 - 3y^2 + z^2) \) gives the components:1. \( \frac{\partial V}{\partial x} = 2Ax \)2. \( \frac{\partial V}{\partial y} = -6Ay \)3. \( \frac{\partial V}{\partial z} = 2Az \)Thus, \[ \vec{E} = - (2Ax \hat{i} - 6Ay \hat{j} + 2Az \hat{k}) = (-2Ax \hat{i} + 6Ay \hat{j} - 2Az \hat{k}) \].

Step 2: Find Constant A from Work Done

Work done by the electric field \( W = q \Delta V \), where \( q \) is the charge and \( \Delta V \) is the change in potential. For a test charge of \( 1.50 \mu C = 1.50 \times 10^{-6} C \) moving from \((0,0,0.250)\) to \((0,0,0)\), \( W = 6.00 \times 10^{-5} J \).The potential difference \( \Delta V = V(0,0,0) - V(0,0,0.250) = 0 - A(0.250)^2 \). Therefore,\[ 6.00 \times 10^{-5} = 1.50 \times 10^{-6} \times (-A(0.250)^2) \]\[ 6.00 \times 10^{-5} = -A(1.50 \times 10^{-6})(0.0625) \]Solving for \( A \):\[ A = -\frac{6.00 \times 10^{-5}}{1.50 \times 10^{-6} \times 0.0625} = -640 \]

Step 3: Calculate Electric Field at Specific Point

Using the expression for \( \vec{E} \) from Step 1 and substituting \( (x, y, z) = (0, 0, 0.250) \) with \( A = -640 \):\[ \vec{E} = (-2(-640)(0) \hat{i} + 6(-640)(0) \hat{j} - 2(-640)(0.250) \hat{k}) \]\[ \vec{E} = (0 \hat{i} + 0 \hat{j} + 320 \hat{k}) = 320 \hat{k} \].

Step 4: Show Equipotential Contours are Circles

In planes parallel to the xz-plane (constant y), for a fixed potential \( V \), the equation is \( x^2 + z^2 = \frac{V + 3A y^2}{A} \). This matches the form of a circle in the xz-plane with radius \( r \) given by:\[ r = \sqrt{\frac{V + 3Ay^2}{A}} \].Thus, equipotential contours are circles.

Step 5: Radius of Equipotential Contour

For \( V = 1280 V \) and \( y = 2 \) m, using the equation \( x^2 + z^2 = \frac{V + 3A y^2}{A} \) from Step 4:Given \( A = -640 \):\[ x^2 + z^2 = \frac{1280 + 3(-640)(2)^2}{-640} \]\[ x^2 + z^2 = \frac{1280 - 7680}{-640} \]\[ x^2 + z^2 = 10 \]Thus, the radius is \( r = \sqrt{10} \) meters.

Key Concepts

Electric Field

The electric field is a vector field surrounding electric charges, representing the force experienced by a positive test charge placed in the field. In electrostatics, the electric field is related to the electric potential by the negative gradient of the potential. This relationship can be expressed mathematically as:
\[ \vec{E} = - abla V \]
Where \( abla \) is the gradient operator in Cartesian coordinates. Calculating the gradient involves taking the partial derivatives of the potential with respect to spatial coordinates. For the potential \( V(x, y, z) = A(x^2 - 3y^2 + z^2) \), the electric field components are derived as follows:

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

This results in the electric field \( \vec{E} = (-2Ax \hat{i} + 6Ay \hat{j} - 2Az \hat{k}) \). The negative sign indicates the direction of the electric force is opposite to the increase in potential.

Electric Potential

Electric potential, often denoted as \( V \), represents the potential energy per unit charge at a point in space due to an electric field. It provides a scalar measure of the work done in bringing a unit positive charge from infinity to that point without any acceleration. Given its scalar nature, the potential is often simpler to work with compared to vector fields.
In this particular exercise, the electric potential is defined as \( V(x, y, z) = A(x^2 - 3y^2 + z^2) \). This expression shows how potential varies with position in a Cartesian coordinate system. It essentially reveals how energy landscape varies due to the presence of charges. Different values of \( A \) help in determining potential at different points, critical for understanding how fields affect charged particles.
The relationship of potential to electric field is crucial, as shown by the gradient relationship discussed earlier. Electric field lines generally move from high potential regions to low potential regions. Thus, analyzing potential can give critical insights into the behavior of electric fields.

Equipotential Surfaces

Equipotential surfaces or contours are regions in space where the electric potential is constant. In these regions, since potential does not vary, no work is needed to move a charge, making these surfaces useful for visualizing electric fields.
This exercise investigates how potential contours behave for planes parallel to the \( xz \)-plane. By keeping \( y \) constant, the potential simplifies to a function of \( x \) and \( z \), resulting in an equation \( x^2 + z^2 = \frac{V + 3Ay^2}{A} \). This equation describes a circle in the \( xz \)-plane. Therefore, equipotential surfaces in such planes appear as circles with a radius given by:

• \( r = \sqrt{\frac{V + 3Ay^2}{A}} \)

This concept helps visualize how the potential landscape looks and reinforces the idea that the electric field is perpendicular to these surfaces.

Work Done by Electric Field

The work done by an electric field in moving a charge from one point to another indicates how much energy is transferred to or from a charge. It's computed using the expression:
\[ W = q \Delta V \]
where \( q \) is the charge, and \( \Delta V \) is the change in electric potential between two points. In the given scenario, a charge of \( 1.50 \mu C \) moves from \((0, 0, 0.250)\) to the origin, performing a work of \( 6.00 \times 10^{-5} \text{ J} \).

• The change in potential \( \Delta V \) is calculated from potential expressions at both points.
• The constant \( A \) is determined by rearranging and solving the work expression.

This analysis is not just academic; it is practical because knowing how much work the field does gives insight into energy changes and system dynamics. It highlights the connection between field, potential, and movement of charges.