Analyzing the Sphere Image Charge Problem

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Homework Help Overview

The discussion revolves around a problem involving a charge placed near a grounded conducting sphere, specifically focusing on finding the electric potential outside the sphere. The subject area includes electrostatics and the method of image charges.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the role of the angle theta in the problem and its relation to the distances involved. There are inquiries about how to analytically determine the values of Q and b, with some suggesting to eliminate theta from the equations. The concept of multiple solutions is also explored, particularly regarding the conditions for zero potential at the sphere's surface.

Discussion Status

The discussion is active, with participants sharing their attempts and questioning the validity of certain equations. Some guidance has been offered regarding the implications of squaring equations and the existence of trivial solutions. There is an ongoing exploration of the constraints imposed by the problem and the relationships between the variables.

Contextual Notes

Participants note that R, d, and q are given parameters, while b and Q are outputs derived from the equations. There is a recognition that the original equation must be solved to fully understand the relationships and constraints involved.

Kaguro
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Homework Statement


A charge q is kept at distance d from center of grounded conducting sphere of radius R. Find V everywhere outside.

Homework Equations


V = k*q/r
Cosine Law.

The Attempt at a Solution


1552928030417769769258.jpg

[/B]
What do I do with the theta?

And how would I analytically find out what Q and b are?

Sorry, I don't know how to use LaTeX...
 

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Kaguro said:

Homework Statement


A charge q is kept at distance d from center of grounded conducting sphere of radius R. Find V everywhere outside.

Homework Equations


V = k*q/r
Cosine Law.

The Attempt at a Solution


View attachment 240483
[/B]
What do I do with the theta?

And how would I analytically find out what Q and b are?

Sorry, I don't know how to use LaTeX...
Just multiply out as necessary to get rid of the square roots and fractions and see what values make theta disappear,
 
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Kaguro said:
What do I do with the theta?
Theta is one of the two independent variables for P. The distance betwwen P and Q and P and q are functions of theta as well as r.
And how would I analytically find out what Q and b are?
Post 2 tells you that. Or you can cheat by googling :smile: It's widely disseminated.
Note that you find b and Q by eliminating theta but the expression for the potential itself must include theta.
 
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haruspex said:
Just multiply out as necessary to get rid of the square roots and fractions and see what values make theta disappear,

Ok!

So I solved it and got this:

b=Q2*d/q2

So this is a parabola! And all points on this must satisfy the condition for 0 potential at surface of sphere... isn't it?

Griffith's has taken one solution
Q= qR/d
b= R2/d

But I can take other solutions too, can't I?
 
Kaguro said:
Ok!

So I solved it and got this:

b=Q2*d/q2

So this is a parabola! And all points on this must satisfy the condition for 0 potential at surface of sphere... isn't it?

Griffith's has taken one solution
Q= qR/d
b= R2/d

But I can take other solutions too, can't I?
In fact, you'd better since the equation for Q is wrong! (Maybe typo.)
 
rude man said:
In fact, you'd better since the equation for Q is wrong! (Maybe typo.)
Ummm... why?

15529733540540.3067998459034783.jpg


Are you talking about minus sign?
 

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Kaguro said:
Are you talking about minus sign?
Yes.
Bear in mind that when you square to get rid of a square root you double the number of solutions of your equations.
Also, there is a trivial solution, Q=-q, b=d, which gives a zero potential everywhere.
Combining those, you will have four solutions, only one of which is nontrivial and satisfies the actual problem.
 
haruspex said:
Yes.
Bear in mind that when you square to get rid of a square root you double the number of solutions of your equations.
Also, there is a trivial solution, Q=-q, b=d, which gives a zero potential everywhere.
Combining those, you will have four solutions, only one of which is nontrivial and satisfies the actual problem.
Just 4 solutions?

That means my parabola equation is not the only constraint?:oldeek:
 
Kaguro said:
Just 4 solutions?

That means my parabola equation is not the only constraint?:oldeek:
Not sure what exactly your parabola equation is. R, d and q are all givens; only b and Q are outputs.
 
  • #10
haruspex said:
Not sure what exactly your parabola equation is. R, d and q are all givens; only b and Q are outputs.

The one which I got when Inequated the theta containing terms:
b= (Q^2)*d/q^2

Is this not enough a constraint?
 
  • #11
Kaguro said:
The one which I got when Inequated the theta containing terms:
b= (Q^2)*d/q^2

Is this not enough a constraint?
What about the terms that do not involve theta? They have to balance too.
 
  • #12
haruspex said:
What about the terms that do not involve theta? They have to balance too.

Ok so I need to solve the original equation...

I feel better now that I know where the solution comes from.

Thank you guys!:smile::smile:
 

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