What is the Solution to the Equipotential Surface Problem?

[gracy]

Homework Statement

[/B]

Homework Equations

$\vec{E}$=$\frac{∂V}{∂r}$

The Attempt at a Solution

I have provided both problem and solution(almost)but the problem is I did not understand the solution.First of all
I did not understand the question what we are told to determine?
I locate the charges as follows

 

[andrewkirk]

Did you write the red part? It seems pretty sensible to me, so if you wrote it I think you've finished.

If it's a supplied solution and you don't understand it, can you explain what it is that troubles you about it? That would make it easier to help.

I don't know if this is relevant to your concern but I'll point out anyway that the surface they've asked you to identify is the only flat equipotential surface. All the others curve away from the flat surface. So the problem in general isn't as trivial as it might seem (if it's the apparent triviality that worries you).

 

[gracy]

If it's a supplied solution

Yes.

 

[gracy]

can you explain what it is that troubles you about it?

I did not understand the question what we are told to determine?

I mean the question mentioned two charges located at rest and then suddenly it (the question)points out at equipotential surface.I don't understand from where does this equipotential surface come?

 

[haruspex]

I mean the question mentioned two charges located at rest and then suddenly it (the question)points out at equipotential surface.I don't understand from where does this equipotential surface come from?

Do you understand what an equipotential surface is?

 

[azizlwl]

The flux is radiating radially outward for positive charges and inward if negative. For stand alone charge, any circle around it will be equipotential. If you introduce another charges, every point is determine by the resultant potential.

 

[gracy]

Do you understand what an equipotential surface is?

Equipotential surfaces are surfaces of constant scalar potential.

 

[haruspex]

Equipotential surfaces are surfaces of constant scalar potential.

Right. So, in the configuration of charges given, you are trying to find the locus of all the points at some particular potential. In general that locus will be a surface, but it is here given as a 2D problem, so you are just looking for a closed curve in the plane. You are told a point that is on it (which effectively determines what the constant potential is).
Suppose the point (x, y) is on the equipotential. Write out the expression for the potential at that point.

 

[gracy]

Right?

 

[haruspex]

View attachment 92380
Right?

Yes.

 

[gracy]

Write out the expression for the potential at that point.

How?

 

[gracy]

How?

Potential of the equipotential surface shown(by me in post #9)
$V$=$\frac{q}{4πεo r}$

Potential due to +1.0 C=$\frac{1}{4πεo 0.5}$

Potential due to -1.0 C=$\frac{-1}{4πεo 0.5}$

Net $V$ at the point =0

Hence this potential is constant for the entire surface
Right?

 

[gracy]

equipotential surface(especially when it is 2d) is just made up of many points at which potential is of same value.Right?

 

[gracy]

What I don't understand is we are given only two points how can we determine that the entire surface should be represented by straight line?beyond those two points ,surface can take any route/shape.

 

[gracy]

What I don't understand is we are given only two points how can we determine that the entire surface should be represented by straight line?beyond those two points ,surface can take any route/shape.

surface they've asked you to identify is the only flat equipotential surface. All the others curve away from the flat surface

I think this somehow answers my question.But problem is I did not understand what @andrewkirk has written.

 

[Doc Al]

Potential of the equipotential surface shown(by me in post #9)
$V$=$\frac{q}{4πεo r}$

Potential due to +1.0 C=$\frac{1}{4πεo 0.5}$

Potential due to -1.0 C=$\frac{-1}{4πεo 0.5}$

Net $V$ at the point =0

You've calculated the potential at a single point.

Hence this potential is constant for the entire surface
Right?

The trick is to realize that, due to symmetry, the potential at any point on the plane y = 1.5 will be zero.

 

[andrewkirk]

What I don't understand is we are given only two points how can we determine that the entire surface should be represented by straight line?beyond those two points ,surface can take any route/shape.

Each equipotential surface is a surface in 3D space, which corresponds to a line on the 2D graph, defined by the equation 'potential = P'. Each value of P gives a different line. The line they have asked you to find is the one where P=0.

The potential at a point $(x,y)$ is calculated as the sum of the potentials at that point created by each of the two charges, ie
$\frac{1}{4\pi\epsilon_0}\bigg(\frac{1}{d((x,y)(1,1))}-\frac{1}{d((x,y),(1,2))}\bigg)$ where $d((x,y),(u,v))$ is the distance between points $(x,y)$ and $(u,v)$.

For P=0, this is a straight line that bisects the two charges. For P>0 it will be a downwards curved line, reflectively symmetric in the line x=1, with its maximum at x=1 at a point closer to the lower charge [Or it may be the other way around. I haven't bothered to work out which way the signs go because it doesn't matter for the purpose of this exercise] For any given potential value P, the equipotential line will be the line satisfying the equation

$\frac{1}{4\pi\epsilon_0}\bigg(\frac{1}{d((x,y)(1,1))}-\frac{1}{d((x,y),(1,2))}\bigg)=P$

 

[gracy]

For P=0, this is a straight line that bisects the two charges. For P>0 it will be a downwards curved line

How do you know?

by the equation 'potential = P

What's that?

 

[haruspex]

What I don't understand is we are given only two points how can we determine that the entire surface should be represented by straight line?beyond those two points ,surface can take any route/shape.

You would not need to be given any points if you were just given the potential, V. You could then find all the points where the potential is equal to that value.
At the point (x, y), what is the potential due to the charge at (1, 1)? What is the potential due to the charge at (1,2)? Write the equation that the sum equals V.
Instead of being given the value V, you are given two points where the value is V. That allows you to determine V.

 

[gracy]

You would not need to be given any points if you were just given the potential,

Instead of being given the value V,

But I nowhere wrote that we have been given potential V

 

[haruspex]

But I nowhere wrote that we have been given potential V

You have effectively been given the potential. You want the equipotential that passes through (1, 1.5). So find the potential at that point.

 

[gracy]

So find the potential at that point.

I found that.It came out to be Zero.

 

[gracy]

Each value of P gives a different line.

And line is always straight.
Definition of a line:A Line is a straight path that is endless in both directions.
And hence it can not take any route/shape.
Right?

 

[haruspex]

I found that.It came out to be Zero.

Right, so you are looking for the equation relating x and y given that the potential at (x,y) is zero.
What is the general expression for the potential at (x,y) from those two charges?

 

[haruspex]

And line is always straight.
Definition of a line:A Line is a straight path that is endless in both directions.
And hence it can not take any route/shape.
Right?

I often draw lines that are not straight. They're still lines. Ok, so call them curves.

 

[gracy]

Is definition wrong?

 

[haruspex]

I often draw lines that are not straight. They're still lines.

Is definition wrong?

Many words in mathematics and physics have slightly different meanings according to context. If you are a topologist you have to distinguish between round circles and other 'circles'. In the context of formal Euclidean geometry, yes lines and line segments are straight. Not so in projective geometry or topology. Lines of flux are generally not straight.

 

[gracy]

And line is always straight.
Definition of a line:A Line is a straight path that is endless in both directions.
And hence it can not take any route/shape.
Right?

yes lines and line segments are straight.

Here my reasoning is correct,right?[/QUOTE]

 

[haruspex]

Here my reasoning is correct,right?

We're not discussing formal Euclidean geometry here. The context is physics. Stop being so pedantic. Andrew's post discussed curved and straight lines, quite explicitly, so it should have been obvious he was not taking 'line' to mean a straight line.

 

[BvU]

Equipotential surface (in this case equipotential lines in the x,y plane -- where line can be curved or straight) is the collection of points with the same potential. You have been given two points for which you calculate the potential to be zero. Other points for which the potential is also zero will have coordinates that satisfy andrewkirks last equation in his post #17 with P = 0.

When P = 0, you can reduce that equation to $\ \ d(x,y)(1,1)=d(x,y)(1,2)\ \$ which is the case for all points with y = 1.5.

y=1.5 is the equation of a straight line .

No fancy math; easy exercise.

 

[J Hann]

It might help if you use the definition of the electric potential at a point:
"The work done in bringing a unit positive test charge from infinity to that point".
What path or paths (if you choose to use a plane instead of a straight line) , in this case,
would the test charge need to travel so that no work is done in moving the test charge?

 

[gracy]

I think I have understood this question properly that's why I want to show my understanding here

Each equipotential surface is a surface in 3D space, which corresponds to a line on the 2D graph where line can be curved or straight(That we will have to figure out).All the points on it will have same potential .With the two points given,I found out that same potential is zero then I had to find out the points of the equipotenial surface.$\frac{1}{4\pi\epsilon_0}\bigg(\frac{1}{\sqrt{(x-1)^2+(y-1)^2}}-\frac{1}{\sqrt{(x-1)^2+(y-2)^2}}\bigg)$=PThis is the equation for an arbitrary equipotential surface - If we put in the value of P, we will get the equipotential surface whose value of potential on it is equal to P

By solving that I got $y$=$1.5$

in a 3D Cartesian coordinates it is a collection of all points (x,1.5,z) where the values of x and z are arbitrary, any possible values from −∞ to +∞.

Hence in this case the equipotential will be represented by straight line

 

[BvU]

By solving that I got $y=1.5$

By solving that for P = 0

in a 3D Cartesian coordinates it is a collection of all points (x,1.5,z) where the values of x and z are arbitrary, any possible values from −∞ to +∞. Hence in this case the equipotential will be represented by straight line

The intersection of this equipotential plane with the plane z = 0 is a straight line parallel to the x-axis

In short: your understanding is OK !