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 \(1.50-\mu \mathrm{C} \quad\) 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 \(\mathrm{m} ) .\) (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

(a) \(-A(2x)\hat{i} + A(6y)\hat{j} - A(2z)\hat{k}\). (b) \(A = 640 \text{ V/m}^2\). (c) \(-320 \hat{k} \text{ V/m}\). (d) Circles in \(xz\)-planes. (e) Radius is \(\sqrt{2} \text{ m}\).

Step 1: Find the Gradient of the Potential

The electric field \(\vec{E}\) is given by the negative gradient of the electric potential \(V\). Thus, we have:\[\vec{E} = - abla V = - \left(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k}\right)\].Given, \(V(x, y, z) = A(x^2 - 3y^2 + z^2)\), we compute:For \(\frac{\partial V}{\partial x}\):\[\frac{\partial V}{\partial x} = A(2x)\].For \(\frac{\partial V}{\partial y}\):\[\frac{\partial V}{\partial y} = A(-6y)\].For \(\frac{\partial V}{\partial z}\):\[\frac{\partial V}{\partial z} = A(2z)\].Thus, \(abla V = A(2x)\hat{i} - A(6y)\hat{j} + A(2z)\hat{k}\), and the electric field is:\[\vec{E} = -A(2x)\hat{i} + A(6y)\hat{j} - A(2z)\hat{k}\].

Step 2: Calculate Work Done and Find Constant A

The work done \(W\) by the electric field is given by:\[W = -q\Delta V\],where \(q = 1.50 \times 10^{-6} \text{ C}\) and \(\Delta V = V(0,0,0) - V(0,0,0.25)\).Evaluate potentials:\[V(0,0,0) = A(0^2 - 3 \times 0^2 + 0^2) = 0\],\[V(0,0,0.25) = A(0^2 - 3 \times 0^2 + (0.25)^2) = 0.0625A\].So, \(\Delta V = 0 - 0.0625A = -0.0625A\).Thus:\[6.00 \times 10^{-5} = -1.50 \times 10^{-6} \times (-0.0625A)\].Solving gives:\[A = \frac{6.00 \times 10^{-5}}{9.375 \times 10^{-8}} = 640\text{ V/m}^2\].

Step 3: Determine the Electric Field at Specific Point

Using the expression for \(\vec{E}\) derived earlier, and substituting the point \( (0, 0, 0.25) \), we have:\[ \vec{E} = -A(2 \times 0)\hat{i} + A(6 \times 0)\hat{j} - A(2 \times 0.25)\hat{k} \]\[ \vec{E} = -A(0.5)\hat{k} = -320 \hat{k} \text{ V/m} \].

Step 4: Analyze Equipotential Contours

Equipotential contours imply that \(V = \text{constant}\). In the \(xz\)-plane, \(y = \text{constant}\). Thus, \[A(x^2 - 3(\text{constant})^2 + z^2) = \text{constant}\Rightarrow A(x^2 + z^2) = \text{new constant}\].This represents a circle in the \(xz\)-plane centered at origin with radius \(\sqrt{\frac{\text{new constant}}{A}}\). This satisfies the condition for \(xz\) planes.

Step 5: Calculate Radius for Given Potential

Given the potential \(V = 1280\text{ V}\) and \(y = 2.00\text{ m}\), we have:\[1280 = 640(x^2 - 3 \times 2^2 + z^2)\]. Thus,\[2 = x^2 + z^2\].The radius of the equipotential contour is:\[\sqrt{2}\text{ m}\].

Key Concepts

Electric Potential

Electric potential, also known as voltage, is a measure of the potential energy per unit charge at a point in an electric field. It describes how much work is needed to move a charge within the field. The expression for electric potential is often given in a function of position, such as the one in the problem: \[V(x, y, z) = A(x^2 - 3y^2 + z^2)\] where \(A\) is a constant that scales the potential. Understanding electric potential helps us comprehend how an electric charge interacts with the space around it, how much energy it possesses, and how it might move under the influence of an electric field. This potential can vary depending on the configuration and location in space, which could include different surfaces or volumes.

Gradient in Physics

In physics, the gradient is a crucial concept that relates to how a quantity changes as you move through space. Specifically, the gradient of a scalar field, like electric potential, points in the direction of the greatest rate of increase of that field. For an electric potential \(V(x, y, z)\), the gradient reveals the rate of change of this potential in the \(x\), \(y\), and \(z\)-directions. The electric field \(\vec{E}\) at any point can be determined by the negative of this gradient: \[\vec{E} = - abla V = - \left(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k}\right)\] This negative gradient signifies the direction and magnitude at which a positive test charge would accelerate in the field, helping naturally align with the concept of force fields where charges experience a push or pull.

Work Done by Electric Field

Work done by the electric field involves moving a charge within the field and is crucial for understanding energy transformations. The work \(W\) carried out by the field in moving a charge from one point to another is linked to the change in electric potential, expressed as: \[W = -q \Delta V\] where \(q\) is the charge and \(\Delta V\) is the change in potential between the initial and final points. In our problem, a charge is moved, resulting in a specific amount of work, which allows us to calculate the constant \(A\). Understanding this principle helps explain how energy is transferred in an electric field and is essential in scenarios like electric circuits and electrostatic phenomena.

Equipotential Contours

Equipotential contours refer to lines or surfaces of constant potential. In simple terms, these are paths where the electric potential remains unchanged. Imagine topographical maps where the lines represent points of equal height; similarly, equipotential contours show regions of equal electric potential. If a plane is parallel to the \(xz\)-plane, and considering \(y = \text{constant}\), the potential simplifies to a circular expression: \[A(x^2 + z^2) = \text{constant}\] This narration tells us that these planes create circular contours. Their nature helps in visualizing how electric fields distribute in space, and how a charge would need to exert no work to move along these contours. Plus, for a given potential value and specific \(y\)-value, the circular radius can be calculated, revealing the size and shape of equipotential areas. This is a pivotal concept in electromagnetism, aiding in the design and understanding of electrical devices and naturally occurring electric patterns.