Problem 41 - Calculus Early Transcendentals

Question

The electric potential function for two positive charges, one at (0,1) with twice the strength as the charge at \((0,-1),\) is given by $$\varphi(x, y)=\frac{2}{\sqrt{x^{2}+(y-1)^{2}}}+\frac{1}{\sqrt{x^{2}+(y+1)^{2}}}.$$ a. Graph the electric potential using the window \([-5,5] \times[-5,5] \times[0,10].\) b. For what values of \(x\) and \(y\) is the potential \(\varphi\) defined? c. Is the electric potential greater at (3,2) or (2,3)\(?\) d. Describe how the electric potential varies along the line \(y=x.\)

Step-by-Step Solution

Answer

Explain your answer with proper calculations. Answer: The electric potential is greater at point (3,2). We can determine this by calculating the electric potential at both points using the given function: \(\varphi(3, 2)=\frac{2}{\sqrt{3^{2}+(2-1)^{2}}}+\frac{1}{\sqrt{3^{2}+(2+1)^{2}}}=\frac{2}{\sqrt{10}}+\frac{1}{\sqrt{18}}.\) \(\varphi(2, 3)=\frac{2}{\sqrt{2^{2}+(3-1)^{2}}}+\frac{1}{\sqrt{2^{2}+(3+1)^{2}}}=\frac{2}{\sqrt{8}}+\frac{1}{\sqrt{20}}.\) Comparing the values, we find that: \(\frac{2}{\sqrt{10}}+\frac{1}{\sqrt{18}}>\frac{2}{\sqrt{8}}+\frac{1}{\sqrt{20}}.\) Therefore, the electric potential is greater at point (3, 2).

Step 1: a. Graph the electric potential function

To graph the electric potential function given by \(\varphi(x, y)=\frac{2}{\sqrt{x^{2}+(y-1)^{2}}}+\frac{1}{\sqrt{x^{2}+(y+1)^{2}}}\), use graphing software (such as Desmos, GeoGebra, or a graphing calculator) to plot the function within the given window: \([-5,5]\times[-5,5]\times[0,10].\) Here, we cannot draw the graph in plaintext but as a teacher, you can draw it using graphing software and show it to your students.

Step 2: b. For what values of \(x\) and \(y\) is the potential \(\varphi\) defined?

The electric potential function will be undefined when the denominators equal zero. We need to find the values of \(x\) and \(y\) for which this occurs: \(\sqrt{x^{2}+(y-1)^{2}}=0\) and \(\sqrt{x^{2}+(y+1)^{2}}=0.\) Both expressions become zero only when \(x=0\) and \(y=1\) for the first equation, and \(x=0\) and \(y=-1\) for the second equation. These points correspond to the positions of the charges. So, the potential \(\varphi\) is defined everywhere except at \((0, 1)\) and \((0, -1).\)

Step 3: c. Is the electric potential greater at (3,2) or (2,3)?

To compare the electric potential at points (3,2) and (2,3), we can calculate the value of \(\varphi(x, y)\) at both points: \(\varphi(3, 2)=\frac{2}{\sqrt{3^{2}+(2-1)^{2}}}+\frac{1}{\sqrt{3^{2}+(2+1)^{2}}}=\frac{2}{\sqrt{10}}+\frac{1}{\sqrt{18}}.\) \(\varphi(2, 3)=\frac{2}{\sqrt{2^{2}+(3-1)^{2}}}+\frac{1}{\sqrt{2^{2}+(3+1)^{2}}}=\frac{2}{\sqrt{8}}+\frac{1}{\sqrt{20}}.\) We can now compare the values: \(\frac{2}{\sqrt{10}}+\frac{1}{\sqrt{18}}>\frac{2}{\sqrt{8}}+\frac{1}{\sqrt{20}}.\) Therefore, the electric potential is greater at point (3, 2).

Step 4: d. Describe how the electric potential varies along the line \(y=x.\)

To examine the behavior of the electric potential along the line \(y=x\), we can plug \(x\) into the function for \(y\): \(\varphi(x, x)=\frac{2}{\sqrt{x^{2}+(x-1)^{2}}}+\frac{1}{\sqrt{x^{2}+(x+1)^{2}}}.\) Simplifying the expression: \(\varphi(x, x)=\frac{2}{\sqrt{2x^{2}-2x+1}}+\frac{1}{\sqrt{2x^{2}+2x+1}}.\) Notice that when the value of \(x\) increases (in either positive or negative direction) both denominators become larger. As a result, both terms in the expression decrease in value when \(x\) moves away from the charges. Thus, along the line \(y=x\), the electric potential decreases as we move away from the charges and increases as we get closer to the charges.

Key Concepts

Graphing Software

Visualizing mathematical functions can often be complex without the right tools. Graphing software, such as Desmos and GeoGebra, enables us to graph a variety of functions easily. These tools allow you to input the equation for the electric potential function, \[ \varphi(x, y)=\frac{2}{\sqrt{x^{2}+(y-1)^{2}}}+\frac{1}{\sqrt{x^{2}+(y+1)^{2}}} \] within a specific domain, such as the window \([-5,5]\times[-5,5]\times[0,10]\). By visualizing this function, you can better understand how electric potential behaves in space. Here's how graphing software can help:

• Adjust the viewing window to explore specific regions of the function.
• Zoom in and out to see more detail or capture a broader perspective.
• Use sliders to dynamically change variables and see their effect on the graph instantly.
• Overlay multiple graphs to compare different functions or parameters easily.

Such tools are essential in physics and other sciences to provide a clearer, more interactive understanding of complex equations.

Domain of a Function

The domain of a function includes all the possible input values (x, y in two dimensions) for which the function is defined. With the electric potential function given by \(\varphi(x, y)=\frac{2}{\sqrt{x^{2}+(y-1)^{2}}}+\frac{1}{\sqrt{x^{2}+(y+1)^{2}}}\), we must consider where the function is undefined.
This function involves square roots in the denominator. Therefore, it becomes undefined where these denominators equal zero:

• \(\sqrt{x^{2}+(y-1)^{2}}=0\) at the point \((0, 1)\).
• \(\sqrt{x^{2}+(y+1)^{2}}=0\) at the point \((0, -1)\).

These points are where the charges themselves are located, leading to undefined potentials at those spots. Outside of these two points, the function is defined across the entire plane, allowing you to study the influence of these charges on points throughout the space.

Comparing Values

To determine where the electric potential is greater, we calculate its value at different points. Comparisons involve simple substitution into the function. Let's explore the potential at points (3, 2) and (2, 3):
For \(\varphi(3, 2): \frac{2}{\sqrt{3^{2}+(2-1)^{2}}}+\frac{1}{\sqrt{3^{2}+(2+1)^{2}}}=\frac{2}{\sqrt{10}}+\frac{1}{\sqrt{18}}.\)
For \(\varphi(2, 3): \frac{2}{\sqrt{2^{2}+(3-1)^{2}}}+\frac{1}{\sqrt{2^{2}+(3+1)^{2}}}=\frac{2}{\sqrt{8}}+\frac{1}{\sqrt{20}}.\)
By comparing the calculated values, we find that \(\frac{2}{\sqrt{10}}+\frac{1}{\sqrt{18}}\) is greater than \(\frac{2}{\sqrt{8}}+\frac{1}{\sqrt{20}}.\) Thus, the potential is higher at the point (3, 2).

• Higher electric potential implies greater strength or influence of the charges at that point.
• Finding these comparisons helps visualize how potential varies spatially.

Behavior along a Line

Understanding how the electric potential changes along a line can reveal much about the influence of charges. Along the line defined by \(y = x\), we substitute into the potential function:
\[ \varphi(x, x)=\frac{2}{\sqrt{x^{2}+(x-1)^{2}}}+\frac{1}{\sqrt{x^{2}+(x+1)^{2}}} \]
By simplifying, it becomes apparant that as \(x\) moves away from zero (the position of the charges), the expression becomes:\[ increasingly large denominators lead to smaller values of the whole expression, \]indicating decreasing potential.

• The proximity to the charge influences the strength of the potential significantly.
• The further from the charges along \(y = x\), the lower the electric potential becomes.
• Closer to the line y = x, and near the charges, the potential increases sharply due to decreased distance.