V of 2 parallel lines makes cylinders

[bowlbase]

Homework Statement

Expanded from Griffith's (3rd ed) #2.47

Infinitely long wires parallel to x-axis carrying ±λ charge densities intersect the y-axis at ±a.

A) Calculate V(x,y,z) if V(0,0,0)=0
B)Show that the equipotential surfaces are circular
cylinders. Locate the axis and calculate the
radius of the cylinder that is at potential V₀.

Homework Equations

Gauss, -∫Edl
E=-∇V
?x²+y²=r²?

The Attempt at a Solution

I've found the solution to part A easily enough but I'm not sure how to approach B. I know that at the origin the potential is zero along the x-axis but I'm confused by how I should show that these are both circular cylinders.

I'd post my solution for A but it's exactly the same as the solution manual so it's easy enough to find.

Any help would be appreciated.

 

[Bryson]

I'd post my solution for A but it's exactly the same as the solution manual so it's easy enough to find.

Why not look at your solutions manual. . .

 

[bowlbase]

The part b is not part of the original question. Thus, not in the manual.

 

[bowlbase]

hmm, never mind I guess it was in the manual. Sorry thought that he just added it. thanks anyway.

 

[paranormal]

I am intrigued by this problem and its connection to Griffith's textbook. To address part B, we can use the properties of equipotential surfaces and the formula for a circle to show that the equipotential surfaces are indeed circular cylinders.

First, let's define an equipotential surface as a surface where the potential is constant at all points. In this case, we are dealing with a 2-dimensional equipotential surface, since we are only considering the y-axis.

Next, we can use the formula for a circle, x^2 + y^2 = r^2, to represent the shape of the equipotential surfaces. Since we know that the potential is constant along these surfaces, we can say that V(x,y) = V0 for all points on the surface. Plugging this into the equation for a circle, we get (x^2 + y^2) = r^2 = constant.

This means that the equipotential surfaces are circles with a constant radius, which is determined by the potential V0. Since the equipotential surfaces are centered around the y-axis, we can say that the axis of the cylinder is the y-axis.

To calculate the radius of the cylinder, we can use the value of V0 and the known values of x and y at any point on the surface. For example, if we choose the point (x=a, y=0), we can plug these values into the equation (x^2 + y^2) = r^2 = constant to get (a^2 + 0^2) = V0. This means that the radius of the cylinder is √V0.

In conclusion, we have shown that the equipotential surfaces are indeed circular cylinders, with the y-axis as the axis and a radius of √V0. This further illustrates the relationship between electric potential and equipotential surfaces, and how they can help us visualize and understand electric fields.