Question

Use a computer as needed to make plots of the given surfaces and the isothermal or equipotential curves. Try both 3D graphs and contour plots. (a) Given \(\phi=x^{2}-y^{2},\) sketch on one graph the curves \(\phi=4, \phi=1, \phi=0\), \(\phi=-1, \phi=-4 .\) If \(\phi\) is the electrostatic potential, the curves \(\phi=\) const. are equipotentials, and the electric field is given by \(\mathbf{E}=-\nabla \phi\). If \(\phi\) is temperature, the curves \(\phi=\) const. are isothermals and \(\nabla \phi\) is the temperature gradient; heat flows in the direction \(-\nabla \phi\). (b) Find and draw on your sketch the vectors \(-\nabla \phi\) at the points \((x, y)=(\pm 1,\pm 1)\), \((0,\pm 2),(\pm 2,0) .\) Then, remembering that \(\nabla \phi\) is perpendicular to \(\phi=\) const., sketch, without computation, several curves along which heat would flow [see(a)].

Short answer

Plot \(x^2 - y^2 = \) for \( \phi = 4, 1, 0, -1, -4 \), and calculate \( -abla \phi \) at given points like (1,1) to get vectors. Finally, sketch heat flow curves perpendicular to equipotentials.

Step-by-step solution

Understand the given function and values

The function given is \(\phi = x^{2} - y^{2} \).We need to plot equipotential (or isothermal) curves for \( \phi = 4, \phi = 1, \phi = 0, \phi = -1, \phi = -4 \).These are the curves where the function remains constant.

Rewrite the equations for the given values of \(\phi\)

For \( \phi = 4 \), rewrite \( x^{2} - y^{2} = 4 \).Similarly,for \(\phi = 1 \) rewrite \( x^{2} - y^{2} = 1 \), \(\phi = 0 \), \(x^{2} = y^{2} \),For \( \phi = -1 \) rewrite \( x^{2} - y^{2} = -1 \),and for \(\phi = -4 \), rewrite \( x^{2} - y^{2} = -4 \).

Plot the equipotential curves

Use a graphing tool or software like Desmos, GeoGebra, or any graphing calculator to plot the curves \( x^2 - y^2 = 4 \), \( x^2 - y^2 = 1 \), \( x^2 = y^2 \), and \( x^2 - y^2 = -1, x^2 - y^2 = -4 \) on the same coordinate plane.Mark these curves on a single graph.

Calculate the gradient \(abla \phi\)

The gradient of \(\phi \) is \(abla \phi = \left( \frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y} \right) \).Given \( \phi = x^2 - y^2 \),\begin{align*}\frac{\partial \phi}{\partial x} &= 2x \frac{\partial \phi}{\partial y} &= -2y\end{align*}.Therefore,\(abla \phi = (2x, -2y) \).

Calculate and draw vectors \( -abla \phi \) at given points

Calculate \( -abla \phi \) at the points \( ( \pm 1, \pm 1) \), \( (0, \pm 2) \), \( (\pm 2, 0) \):\(-abla \phi \bigg|_{(1, 1)} = -abla \phi \bigg|_{(1, 1)} = (-2 \times 1, 2 \times 1) = (-2, 2) \)Similarly for \( (1, -1) = (-2, -2) \), \( (-1, 1) = (2, 2) \), \( (-1, -1) = (2, -2) \)For points \((0, 2) = (0, 4) = (0, -4) \) and \( (0, -2) = (0, -4) = (0, 4) \), \((2, 0) = (-4, 0) = (-4, 0) \), and \( (-2, 0) = (4, 0) = (4, 0) \).Plot these vectors on the same graph.

Sketch the heat flow curves

Since the heat flows in the direction of \( -abla \phi \), sketch several curves perpendicular to the equipotential curves that represent the direction of heat flow from higher potential to lower potential without computing any additional values.

Key concepts

Equipotential Curves

Equipotential curves are essential in both electrostatics and thermodynamics for visualizing potential and temperature fields. These curves represent points where the potential (or temperature) is constant. In the given exercise, the function is \(\phi = x^{2} - y^{2} \). When you set \(\phi \) to specific values like 4, 1, 0, -1, and -4, you get a series of hyperbolic curves. Here are the equations for these equipotential lines:
\(\phi=4\) gives \(\x^2 - y^2 = 4\)
\(\phi=1\) gives \(\x^2 - y^2 = 1\)
\(\phi=0\) gives \(\x^2 = y^2\)
\(\phi=-1\) gives \(\x^2 - y^2 = -1\)
\(\phi=-4\) gives \(\x^2 - y^2 = -4\).
You can use graphing tools like Desmos or GeoGebra to plot these curves simultaneously. These tools will help you visualize how the curves change, but all lines where \(\phi\) is constant form these equipotential surfaces.
The importance of these curves lies in their ability to provide a mapped-out view of where the potential energy (in electrostatics) or temperature (in thermodynamics) will remain the same. This tells us a lot about how energy or temperature is distributed across a field.

Temperature Gradient

The temperature gradient is the rate of change of temperature with respect to distance. It's a vector quantity that points in the direction of maximum temperature decrease. When dealing with temperatures, the gradient of \(\phi\) represents this change. Given our function \(\phi=x^{2} - y^{2}\), the gradient \(\abla \phi\) can be computed using partial derivatives:
\(\abla \phi = \left( \frac{\partial \phi}{\partial \x}, \frac{\partial \phi}{\partial \y} \right) = (2\x, -2\y) \).
This means the temperature gradient at any point in the field is directed as \(\ (2\x, -2\y) \). To understand the practical relevance of this, consider that heat flows from regions of higher temperature to regions of lower temperature. This is depicted by the negative gradient: \(\mathbf{E} = -abla \phi\). So, at any point, the heat flow direction is exactly opposite to \(\abla \phi\).
Heat always flows in a perpendicular direction to equipotentials (or isothermals), following the path of steepest descent. By plotting the negative gradients at specific points—such as \(\ ( \pm 1, \pm 1)\), \(\ (0, \pm 2)\) and \(\ ( \pm 2, 0)\)—we visualize this heat flow clearly. Always remember: the temperature gradient gives you not just the rate of temperature change but also the direction in which temperature is changing most rapidly.

Electric Field

In the context of electrostatics, an electric field signifies the force experienced by a unit positive charge. It points from positive to negative potential. Hence, it’s oriented perpendicular to equipotential surfaces. Its direction and magnitude are given by \(\mathbf{E} = -abla \phi\).
This essential relationship tells us that electric fields are derived from the negative gradient of the electric potential. For the given function, the gradient \(\abla \phi \) is \(\ (2\x, -2\y) \). Thus, the electric field is:\
\(abla \phi = - (2\x, -2\y) = (-2\x, 2\y) \).
This shows us that the electric field’s direction is opposite that of the potential gradient. For the points specific in the exercise—\(\ ( \pm 1, \pm 1), (0, \pm 2), (\pm 2, 0) \)—you can calculate \(\mathbf{E}\) as follows:
At \( (1,1) = (-2,2)\)
At \( (1, -1) = (-2,-2)\)
At \( (-1, 1) = (2, 2)\)
At \( (-1,-1) = (2, -2)\)
The field indicates where a charge would move: from high potential regions to low potential regions. By understanding how \(\abla \phi\) and \(\mathbf{E} \) are correlated, it becomes clear that equipotential lines and electric field lines are inherently interconnected concepts. Together, they comprehensively describe how charges and heat distribute themselves in a given potential field.