Analyzing the Sphere Image Charge Problem

[Kaguro]

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

[/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...

 

[haruspex]

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,

 

[rude man]

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 It's widely disseminated.
Note that you find b and Q by eliminating theta but the expression for the potential itself must include theta.

 

[Kaguro]

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?

 

[rude man]

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.)

 

[Kaguro]

In fact, you'd better since the equation for Q is wrong! (Maybe typo.)

Ummm... why?

Are you talking about minus sign?

 

[haruspex]

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.

 

[Kaguro]

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?

 

[haruspex]

Just 4 solutions?

That means my parabola equation is not the only constraint?

Not sure what exactly your parabola equation is. R, d and q are all givens; only b and Q are outputs.

 

[Kaguro]

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?

 

[haruspex]

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.

 

[Kaguro]

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!