Question

If \(V(x, y)\) is the voltage at a point \((x, y)\) in the plane, the level curves of \(V\) are called equipotential curves. Draw the equipotential curves corresponding to \(V=\frac{1}{2}, 1,2,4\) for $$ V(x, y)=\frac{4}{\sqrt{(x-2)^{2}+(y+3)^{2}}} $$

Short answer

The equipotential curves are circles centered at (2, -3) with radii 8, 4, 2, and 1.

Step-by-step solution

Understand Cloud of Equipotential Curves

To find equipotential curves for a given voltage, we need to set each voltage level equal to the function of voltage, which in this case is \(V(x, y) = \frac{4}{\sqrt{(x-2)^2 + (y+3)^2}}\). The equipotential curves will be circles because the voltage function resembles a radial distance function from the point (2, -3).

Set Up Equations for Each Voltage Level

For each given voltage \(V = \frac{1}{2}, 1, 2, 4\), we set \(\frac{4}{\sqrt{(x-2)^2 + (y+3)^2}} = V\). Rearranging gives:- For \(V = \frac{1}{2}\): \(\sqrt{(x-2)^2 + (y+3)^2} = 8\)- For \(V = 1\): \(\sqrt{(x-2)^2 + (y+3)^2} = 4\)- For \(V = 2\): \(\sqrt{(x-2)^2 + (y+3)^2} = 2\)- For \(V = 4\): \(\sqrt{(x-2)^2 + (y+3)^2} = 1\)

Solve for Radii of Circles

Solve the square root equations from Step 2 to find the radius of each circle:- For \(V = \frac{1}{2}\), the radius is 8.- For \(V = 1\), the radius is 4.- For \(V = 2\), the radius is 2.- For \(V = 4\), the radius is 1.

Write Circle Equations

For a circle centered at \((2, -3)\) with radius \(r\), the equation is \((x-2)^2 + (y+3)^2 = r^2\):- For \(V = \frac{1}{2}\), \((x-2)^2 + (y+3)^2 = 64\).- For \(V = 1\), \((x-2)^2 + (y+3)^2 = 16\).- For \(V = 2\), \((x-2)^2 + (y+3)^2 = 4\).- For \(V = 4\), \((x-2)^2 + (y+3)^2 = 1\).

Key concepts

Voltage Function

A voltage function is a mathematical expression that represents the electric potential distribution in a given space. For this specific task, the voltage at a point \((x, y)\) in the plane is represented as \(V(x, y) = \frac{4}{\sqrt{(x-2)^2 + (y+3)^2}}\). This particular function describes how the voltage decreases with increasing distance from the source point addressable at \((2, -3)\).
Understanding the voltage function is crucial as it determines how the equipotential lines, also known as level curves, will extend outwards.
These curves depict areas where the voltage remains constant. Thus, for voltages \(V = \frac{1}{2}, 1, 2, 4\), we set these values equal to the function to find specific locations of constant voltage in space.

Radial Distance Function

The radial distance function is indicative of how far a point lies away from a given central point. In the context of this problem, the radial distance function describes the distance from the point \((2, -3)\) in the 2D plane. The expression \(\sqrt{(x-2)^2 + (y+3)^2}\) serves as the radial distance, measuring how far any point \((x,y)\) is from \((2, -3)\).
This function is pivotal because the original voltage function \(V(x, y) = \frac{4}{\text{radial distance}}\) shows that the voltage levels are governed by this distance.
To find the equipotential curves, we solve this distance function for different voltage levels, translating them into circles around \((2, -3)\). The value of the radial distance gives us the radius of the circles.

Circle Equation

The circle equation is an algebraic way to describe a circle's properties, such as its center and radius. In this exercise, the general form of a circle centered at the point \((2, -3)\) is \((x-2)^2 + (y+3)^2 = r^2\).
For each voltage level, solving the rearranged version of the voltage equation gives us different radii:

• For \(V = \frac{1}{2}\), the radius is found to be \(8\)
• For \(V = 1\), the radius is \(4\)
• For \(V = 2\), the radius is \(2\)
• For \(V = 4\), the radius is \(1\)

Once we identify these radii, each corresponding circle equation can be represented as:

• \((x-2)^2 + (y+3)^2 = 64\)
• \((x-2)^2 + (y+3)^2 = 16\)
• \((x-2)^2 + (y+3)^2 = 4\)
• \((x-2)^2 + (y+3)^2 = 1\)

This circles drawn based on these equations show that as voltage increases, the radius of the circles decreases, confirming the V-voltage-to-distance relationship.

Level Curves

Level curves, also known as equipotential curves in physics, are lines along which a particular function has a constant value. For voltage functions, these curves represent points at the same electric potential. In this exercise, the function \(V(x, y)\) translates these level curves into concentric circles around the center point \((2,-3)\).
Here’s how it works:

• For each given voltage level \(V = \frac{1}{2}, 1, 2, 4\), equipotential curves form at specific radii found earlier.
• The result is a series of circles that shrink as the potential voltage \(V\) increases, clearly showing the radial nature of the voltage field surrounding the point.

Visualizing these curves is helpful in understanding how electric charges interact with their environment.
They give us insight into how the strength of the electric field changes with distance from the center point. These levels are fundamental concepts in electromagnetism, aligning closely with field line representations in physics.