Show that the equipotential lines are circles

In summary: 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...In summary, the conversation discusses a problem involving an electrical potential function in a specific area of space. The first part involves determining the electric field and work performed on a test charge by the electrical force, while the second part involves showing that the equipotential lines in planes parallel to the (yz)-plane are circles and determining the radius of these lines for given values. The key equations used are given and the attempt at a solution is described, including solving for A as a function of W_0, q_
  • #1
Erik P
33
0

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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  • #2
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?
 
  • #3
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.
 
  • #4
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
 
  • #5
ehild said:
What is x on the (y,z) plane?
View attachment 97574
zero? Sorry, not entirely following here.
 
  • #6
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
 
  • #7
ehild said:
yes.
And what is x on the blue plane, parallel with the (y,z) plane?
View attachment 97575
Five, I guess?
 
  • #8
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?
 
  • #9
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:
 

FAQ: Show that the equipotential lines are circles

What are equipotential lines?

Equipotential lines are imaginary lines that connect points on a surface with equal potential. In other words, the potential remains the same at every point on an equipotential line.

Why are equipotential lines important?

Equipotential lines are important because they help us visualize the electric field and understand the behavior of charged particles in an electric field. They also help us determine the direction of the electric field at any given point.

How are equipotential lines related to electric potential?

Equipotential lines are directly related to electric potential. The potential difference between two points is equal to the work done per unit charge in moving a charge between those points. Therefore, points on the same equipotential line have the same potential.

Why are the equipotential lines circular?

The equipotential lines are circular because the electric field lines around a point charge are radial and perpendicular to the equipotential lines. This creates a circular pattern of equipotential lines around a point charge.

Can equipotential lines ever intersect?

No, equipotential lines can never intersect. If they were to intersect, it would imply that two points on the same line have different potentials, which is impossible. Equipotential lines must always be parallel and evenly spaced.

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