Question

The electric potential \(V\) at any point \((x, y)\) is $$V(x, y)=\frac{5}{\sqrt{25+x^{2}+y^{2}}}$$ Sketch the equipotential curves for \(V=\frac{1}{2}, V=\frac{1}{3},\) and \(V=\frac{1}{4}\).

Short answer

The equipotential curves are circles centered at the origin with radii of 10, 15, and 20 units for the potentials \( V=\frac{1}{2} \), \( V=\frac{1}{3} \), and \( V=\frac{1}{4} \), respectively.

Step-by-step solution

Set Up Equations for Each Given Potential

First, take each given potential value and set it equal to the given V(x, y). This gives: \( \frac{5}{\sqrt{25+x^{2}+y^{2}}} = \frac{1}{2} \), \( \frac{5}{\sqrt{25+x^{2}+y^{2}}} = \frac{1}{3} \), \( \frac{5}{\sqrt{25+x^{2}+y^{2}}} = \frac{1}{4} \). Solve each of these for \( \sqrt{x^{2}+y^{2}} \) to find the radius for each equipotential curve.

Solve the Equations

Solving these equations gives: \( \sqrt{x^{2}+y^{2}} = 10 \) for \( V=\frac{1}{2} \), \( \sqrt{x^{2}+y^{2}} = 15 \) for \( V=\frac{1}{3} \), and \( \sqrt{x^{2}+y^{2}} = 20 \) for \( V=\frac{1}{4} \). This means that for each potential, the equipotential curves are circles centered at the origin with radii of 10, 15, and 20, respectively.

Draw the Curves

Finally, draw circles on the xy coordinate plane with the given radii. Remember to draw the circles centered at the origin. The circle with radius 10 corresponds to \( V=\frac{1}{2} \), the circle with radius 15 corresponds to \( V=\frac{1}{3} \), and the circle with radius 20 corresponds to \( V=\frac{1}{4} \).

Key concepts

Electric Potential

Understanding the concept of electric potential is crucial when tackling problems in electrostatics. Think of electric potential, denoted as V, as the electric potential energy per unit charge at a specific point in an electric field. In simpler terms, it's a measure of the work done by an electric field in moving a unit charge from a reference point to the point in question.

In our exercise, the electric potential is given by the equation \( V(x, y) = \frac{5}{\sqrt{25 + x^2 + y^2}} \), where \(x\) and \(y\) are coordinates in space. This equation provides a three-dimensional surface of potential values over the x-y plane; however, we are interested in visualizing the two-dimensional equipotential curves where the potential value is constant.

The relevance of these curves is that they can visually demonstrate how potential varies with position. They also show that no work is done when moving along these lines; that's why they are 'equipotential'. In the given solution, various values for V allow us to plot circles of constant potential, giving a clear insight into the symmetry and distribution of electrical potential around a point.

Sketching Curves

The exercise moves from the abstract concept of electric potential to the more visual task of sketching equipotential curves. Sketching these curves aids in understanding the relationship between electric potential and position. It's like creating a topographic map of gravitational potential if gravity varied in two dimensions like our electric potential does here.

To sketch these curves accurately, one must recognize that electric potential values on an equipotential curve are all the same. In the context of our exercise, the curves are circles because the potential's equation is dependent only on the distance from the origin, not the direction. This isotropy means potential is the same at any point equidistant from the origin, forming a circle with the origin as the center.

By plotting the circles with radii proportional to the potential values given in the problem, students can create a visual representation that solidifies their understanding of how potential varies in space. Each circle on our xy-plane represents a 'slice' of potential values, making it easier to comprehend the continuous nature of electric fields.

Solving Equations

Any problem in physics often comes down to a universal skill: solving equations. This skill is integral when you are presented with a formula for electric potential and asked to determine the points that have specific potential values. In our exercise, solving equations is a two-step process. First, you substitute the known potential values into the general formula for the potential, and then rearrange to solve for the radius, which in our case is represented by the term \(\sqrt{x^2 + y^2}\).

Let's break it down: To find the radius for the circles that represent the equipotential curves, we isolate \(\sqrt{x^2 + y^2}\) on one side by cross-multiplying and simplifying. For instance, when \(V = \frac{1}{2}\), we rearrange to solve for the radius, which gives us \(\sqrt{x^2 + y^2} = 10\). This method is repeated for the other potential values as well.

It's important for students to be adept at rearranging equations and isolating variables, as it's a technique that will reoccur throughout their studies in physics and other sciences. Moreover, understanding how to manipulate equations provides deeper insight into the relationships between quantities in a physical system.