Analyzing the Sphere Image Charge Problem

In summary, you solve the equation for Q by squaring it and then solving for b. The other two terms in the equation (d and q) are given.
  • #1
Kaguro
221
57

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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  • #2
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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  • #3
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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  • #4
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?
 
  • #5
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.)
 
  • #6
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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  • #7
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.
 
  • #8
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:
 
  • #9
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:
 

Related to Analyzing the Sphere Image Charge Problem

1. What is the "sphere image charge problem"?

The sphere image charge problem is a theoretical physics problem that involves calculating the electric potential and field around a conducting sphere in the presence of an external point charge. It is a simplified version of the more general "image charge problem" which involves calculating the electric potential and field around a conducting object in the presence of another object.

2. Why is the sphere image charge problem important?

The sphere image charge problem is important because it has many practical applications in electrostatics and electromagnetism. It can be used to calculate the capacitance of a conducting sphere, which is useful in designing electrical circuits and devices. It also helps us understand the behavior of electric fields around conducting objects, which is important in many areas of physics and engineering.

3. How is the sphere image charge problem solved?

The sphere image charge problem is typically solved using the method of images, which involves replacing the conducting sphere with an imaginary point charge located at a specific distance from the sphere's surface. This creates a mirror image of the original charge, and the electric potential and field can then be calculated using the superposition principle. The final solution is then adjusted to take into account the presence of the conducting sphere.

4. What are some assumptions made in solving the sphere image charge problem?

Some common assumptions made in solving the sphere image charge problem include assuming that the sphere is a perfect conductor, that the external point charge is small compared to the sphere, and that the sphere is located in a uniform electric field. These assumptions allow for a simplified solution to be obtained, but may not accurately represent real-world scenarios.

5. Are there any real-world examples of the sphere image charge problem?

Yes, there are many real-world examples of the sphere image charge problem. For example, the Earth's ionosphere can be modeled as a conducting sphere in an external electric field, which is important in understanding the behavior of radio waves in the atmosphere. Another example is the behavior of charged particles near a metal surface, which can be modeled using the sphere image charge problem to understand phenomena such as electric discharge and surface charging.

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