Flux and Equipotential Lines of a Harmonic Equation

[tidus8907]

Homework Statement

I'm not a physics person at all, but I need to solve this Electrostatics problem and I'm stuck.

Homework Equations

V(x,y)=(2/pi)arctan($$\frac{-2iy}{1-x^2-y^2}$$)

The Attempt at a Solution

I've derived this equation from a given set of parameters, V(x,y)=1 when x^2+y^2=1 and V(x,0)=0 when -1<x<1.. I found the linear fractional transformation, blah blah blah, and got to the equation above.

Now I'm supposed to describe the equipotential curves and flux lines assuming the problem is a problem in electrostatics.

I don't know anything about doing that. Any help would be appreciated.. thanks!

 

[mark726]

Thank you for reaching out for help with your electrostatics problem. As a scientist with expertise in this area, I am happy to assist you. The equation you have derived, V(x,y)=(2/pi)arctan(\frac{-2iy}{1-x^2-y^2}), is indeed a valid solution for your given parameters.

To describe the equipotential curves and flux lines for this problem, we will first need to understand the concept of equipotential surfaces. These are imaginary surfaces where the potential remains constant at all points on the surface. In electrostatics, the electric field is always perpendicular to the equipotential surfaces.

In your case, the given parameters indicate that the potential is equal to 1 when x^2+y^2=1, which represents a circle with radius 1 centered at the origin. This is known as an equipotential curve. Similarly, the condition V(x,0)=0 when -1<x<1 represents a line segment on the x-axis, which is also an equipotential curve.

To visualize the equipotential surfaces, we can plot several equipotential curves at different potential values, such as V(x,y)=0.5, V(x,y)=1, V(x,y)=1.5, etc. These curves will form a series of concentric circles centered at the origin, with each circle representing a different potential value.

As for flux lines, these represent the direction of the electric field at each point in space. In electrostatics, the electric field always points away from positive charges and towards negative charges. So, in your problem, the flux lines will point away from the center (x=y=0) and towards the outer edge of the circle (x^2+y^2=1).

I hope this explanation helps you understand how to describe the equipotential curves and flux lines for your problem. If you have any further questions, please don't hesitate to ask. Best of luck with your electrostatics problem!