Show that the equipotential lines are circles

AI Thread Summary
The electrical potential is defined as V(x,y,z) = A(2x^2 - 3y^2 - 3z^2), where A is a constant. The electric field E is calculated as E = 2A(-2x, 3y, 3z), and the work W_0 done on a test charge q_0 moving from (a,0,0) to the origin is W_0 = 2q_0Aa^2. To show that equipotential lines in the (yz)-plane are circles, the potential function is evaluated at x = 1 m, leading to the equation 1000 = A(2 - 3y^2 - 3z^2). This results in a circular equation with a radius of r = sqrt(1/3), confirming the circular nature of the equipotential lines.
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 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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