Show that the equipotential lines are circles

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Homework Help Overview

The discussion revolves around the electrical potential given by the equation V(x,y,z) = A(2x^2 - 3y^2 - 3z^2) and the task of showing that the equipotential lines in planes parallel to the (yz)-plane are circles. Participants are exploring the implications of this potential in relation to the electrical field and the work done on a test charge.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of the electrical field and work done on a test charge. There is uncertainty about the correct approach to demonstrate that equipotential lines are circles, with some questioning the use of voltage equations for point charges. Others suggest focusing on the specific conditions of the problem, such as the values of x, y, and z in the (yz)-plane.

Discussion Status

The discussion has evolved with participants providing insights and clarifications regarding the potential function and its implications. There is an ongoing exploration of how to express the equipotential lines in standard form and determine their characteristics, such as center and radius. Some participants have made progress in formulating the equations needed to describe the circles.

Contextual Notes

Participants are working under the constraints of the problem statement, specifically the values of A, W_0, q_0, and the fixed position of x. There is a noted confusion regarding the application of certain equations and the interpretation of the potential function.

Erik P
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Homework Statement


In a specific area of the space, an electrical potential is given as:

\begin{equation}
V(x,y,z) = A(2x^2 - 3y^2 - 3z^2)
\end{equation}

where A is a constant.

a.) Determine the electrical field E for any given point in the area. A test charge q_0 is moved from the point (x,y,z)=(a,0,0) to origin (0,0,0). Determine the expression for the work W_0 performed on the test charge by the electrical force. Determine A as a function of W_0, q_0, and a.

b.) Show that the equipotential lines in planes parallel to the (yz)-plane are circles. Determine the radius of the equipotentiallines where x=a=1.0 m and V=1000 volts, for q_0 = 1.0 micro columb and W_0 = 2.0 mJ.

Homework Equations


The radius of a circle displaced from origin to a point (h,k):

\begin{equation}
R^2 = (y - h)^2 + (z - k)^2
\end{equation}

General expression for a circle:

\begin{equation}
y^2 + z^2 +Cy + Dz + E = 0
\end{equation}

Voltage due to point charge (really not sure if this is the right one to use, i tried inserting numbers and it gives me the wrong result):

\begin{equation}
V = \frac{q_0}{4\pi\epsilon_0R}
\end{equation}

The Attempt at a Solution



Solved the first part a.)

\begin{equation}
\vec{E} = 2A(-2x\vec{i} + 3y\vec{j} + 3z\vec{k})
\end{equation}

\begin{equation}
W_0 = 2q_0Aa^2
\end{equation}

Part b.) I'm not sure how to tackle, here is what I tried...

\begin{equation}
V = A(2x^2 - 3y^2 - 3z^2) = \frac{q_0}{4\pi\epsilon_0R}
\end{equation}

\begin{equation}
r = \frac{q_0}{A(2x^2 - 3y^2 - 3z^2)4\pi\epsilon_0}
\end{equation}

\begin{equation}
A = \frac{W_0}{2a^2q_0}
\end{equation}

\begin{equation}
r = \frac{2a^2q_0^2}{4W_0A(2x^2 - 3y^2 - 3z^2)\pi\epsilon_0}
\end{equation}

Inserting numbers:

\begin{equation}
r = 4.49
\end{equation}

the answer is supposed to be 0.6 m, so obviously I'm way off. I suspect it has to do, atleast in part with me using the voltage for a point charge. Not sure how to go about this. I also need to show that they are circle, but that's also a little over my head at the moment.
 
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Erik P said:

Homework Statement


In a specific area of the space, an electrical potential is given as:

\begin{equation}
V(x,y,z) = A(2x^2 - 3y^2 - 3z^2)
\end{equation}

where A is a constant.

a.) Determine the electrical field E for any given point in the area. A test charge q_0 is moved from the point (x,y,z)=(a,0,0) to origin (0,0,0). Determine the expression for the work W_0 performed on the test charge by the electrical force. Determine A as a function of W_0, q_0, and a.

b.) Show that the equipotential lines in planes parallel to the (yz)-plane are circles. Determine the radius of the equipotentiallines where x=a=1.0 m and V=1000 volts, for q_0 = 1.0 micro columb and W_0 = 2.0 mJ.

Homework Equations


The radius of a circle displaced from origin to a point (h,k):

\begin{equation}
R^2 = (y - h)^2 + (z - k)^2
\end{equation}

General expression for a circle:

\begin{equation}
y^2 + z^2 +Cy + Dz + E = 0
\end{equation}

Voltage due to point charge (really not sure if this is the right one to use, i tried inserting numbers and it gives me the wrong result):

\begin{equation}
V = \frac{q_0}{4\pi\epsilon_0R}
\end{equation}

The Attempt at a Solution



Solved the first part a.)

\begin{equation}
\vec{E} = 2A(-2x\vec{i} + 3y\vec{j} + 3z\vec{k})
\end{equation}

\begin{equation}
W_0 = 2q_0Aa^2
\end{equation}

Part b.) I'm not sure how to tackle, here is what I tried...

\begin{equation}
V = A(2x^2 - 3y^2 - 3z^2) = \frac{q_0}{4\pi\epsilon_0R}
\end{equation}

You are given the potential finction V(x,y,z). It is not the potential of a point charge. You need the curve in a plane parallel to the (y,z) plane. What do you know about x in that plane?
 
ehild said:
You are given the potential finction V(x,y,z). It is not the potential of a point charge. You need the curve in a plane parallel to the (y,z) plane. What do you know about x in that plane?
I'm not entirely sure to be honest.
 
Erik P said:
I'm not entirely sure to be honest.

What is x on the (y,z) plane?
upload_2016-3-19_17-36-29.png
 
ehild said:
What is x on the (y,z) plane?
View attachment 97574
zero? Sorry, not entirely following here.
 
Erik P said:
zero? Sorry, not entirely following here.
yes.
And what is x on the blue plane, parallel with the (y,z) plane?
upload_2016-3-19_17-45-2.png
 
ehild said:
yes.
And what is x on the blue plane, parallel with the (y,z) plane?
View attachment 97575
Five, I guess?
 
Erik P said:
Five, I guess?
yes. Read the problem, you are given that x=a=1.0 m and V=1000 volts. How the potential function looks like if x= 1.0 m?
 
ehild said:
yes. Read the problem, you are given that x=a=1.0 m and V=1000 volts. How the potential function looks like if x= 1.0 m?
\begin{equation}
V(1,y,z) = A(2 - 3y^2 - 3z^2)
\end{equation}
 
  • #10
Erik P said:
\begin{equation}
V(1,y,z) = A(2 - 3y^2 - 3z^2)
\end{equation}
oooohhhh.. i think i see it, give me a few mins to look at it
 
  • #11
You want the line where V=1000 V. What is its equation?
 
  • #12
ehild said:
You want the line where V=1000 V. What is its equation?
\begin{equation}
1000 = A(2 - 3y^2 - 3z^2)
\end{equation}

\begin{equation}
0 = A(2 - 3y^2 - 3z^2) - 1000
\end{equation}

\begin{equation}
A = \frac{W_0}{2a^2q_0}
\end{equation}

\begin{equation}
0 = \frac{W_0}{2a^2q_0}(2 - 3y^2 - 3z^2) - 1000
\end{equation}
 
  • #13
Erik P said:
\begin{equation}
1000 = A(2 - 3y^2 - 3z^2)
\end{equation}

\begin{equation}
0 = A(2 - 3y^2 - 3z^2) - 1000
\end{equation}

\begin{equation}
A = \frac{W_0}{2a^2q_0}
\end{equation}

\begin{equation}
0 = \frac{W_0}{2a^2q_0}(2 - 3y^2 - 3z^2) - 1000
\end{equation}
Can you bring it to the standard from of a circle? Where is the center? What is the radius? You also know that a=1. ( x=a=1.0 m) Wo and qo are also given.
 
  • #14
ehild said:
Can you bring it to the standard from of a circle? Where is the center? What is the radius? You also know that a=1. ( x=a=1.0 m) Wo and qo are also given.

\begin{equation}
0 = y^2 + z^2 + \frac{2000a^2q_0}{3W_0}-\frac{2}{3} \Rightarrow 0 = y^2 + z^2 - \frac{1}{3}
\end{equation}

Then r should be sqrt(1/3), right? with a center in (0,0)
 
Last edited:
  • #15
Erik P said:
\begin{equation}
0 = y^2 + z^2 + \frac{2000a^2q_0}{3W_0}-\frac{2}{3} \Rightarrow 0 = y^2 + z^2 - \frac{1}{3}
\end{equation}

Then r should be sqrt(1/3), right? with a center in (0,0)
Yes. Good work!
 
  • Like
Likes   Reactions: Erik P
  • #16
ehild said:
Yes. Good work!
Thank you for the help and also for being patient with me :)
 
  • #17
You are welcome :oldsmile:
 

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