Magnitude and direction of the electric dipole

[tsang]

Homework Statement

Three metal balls with unequal radii R1 < R2 < R3 are placed at the vertices of an equilateral triangle, whose sides have length a. The balls are connected by thin metal wires. A positively charged rod is brought into contact with one of the balls and transfers an amount of charge +Q to the system.
(a)Show that the charge Qi on the i-th sphere (i = 1,2,3) is given by
[tex]\frac{Qi}{Q}[/tex]=[tex]\frac{Ri}{R1+R2+R3}[/tex]
(b)Hence show that the potential V at the centre of the triangle is given by
[tex]V=\frac{Q\sqrt{3}}{4\pi\epsilon_0a}[/tex]
(c)What is the magnitude and direction of the electric dipole moment of the system, if two of the balls have equal radii (e.g. R1 = R2)? Hint: electric dipole moments add like vectors.

Homework Equations

[tex]V=\frac{Q}{4\pi\epsilon_0R}[/tex]
[tex]\vec{p}=q\vec{d}[/tex]

The Attempt at a Solution

I kept trying this question,but still have no idea for part (a).
I think I have solved Part B,but I noticed the question uses "hence", does that mean part (b) should use part (a) as part of solution? But I simply subsititute everything to formula [tex]V=\frac{Q}{4\pi\epsilon_0R}[/tex],and I ended up part (b) without using Part (a),is there anything wrong with that?
Part (c) I tried to use formula [tex]\vec{p}=q\vec{d}[/tex],but I don't know what to do next.

Can anyone please help me with some details? Thanks a lot.

 

[Mandeep Deka]

You are close to the answer of the first part

You wrote the expression of potential of a sphere, in terms of the charge it possesses and its radius. Give a thought, what do you think is the consequence of two conductors being joined by a metal wire and a charge be given on one of them??

 

[tsang]

You are close to the answer of the first part

You wrote the expression of potential of a sphere, in terms of the charge it possesses and its radius. Give a thought, what do you think is the consequence of two conductors being joined by a metal wire and a charge be given on one of them??

Thank you,dear friend. But could you please give me more help on details? I truly have no idea about it. I actually don't know which formula to use for Part (a).
About Part (c), do you mean by I should think about torque? since one side would be positive and the other side would be negative?
Please help me. Thank you so much.

 

[Mandeep Deka]

OK, let's get it clear

whenever two bodies are joined by a conducting wire or maybe placed in physical contact to each other, the essential consequence is that they are at EQUAL POTENTIAL!
i.e say for example you have a metal ball with charge Q, and radius 'R', and then you bring it in contact to an uncharged body, say another metal sphere of radius 'r', it will transfer a particular amount of charge to the uncharged sphere, after which you will find that the both the sphere are at same potential. That is what is the basic essence of the question.

In your question, once you give a total charge Q to the system, it will get distributed among the spheres, and at the end their potentials will be same.
Think about it a little more, you have the equation of potential of a sphere in terms of its radius and charge of it, you can do some simple math and get the answer

Hope its clear now!

 

[rwisz]

I would like to commend you for your efforts in attempting to solve this problem. I will provide some guidance to help you understand the concepts and solve the problem.

Part (a):
To solve for the charge on each sphere, we can use the principle of charge conservation, which states that the total charge before and after any interaction must be the same. In this case, the positively charged rod transfers an amount of charge +Q to the system. This means that the total charge on the three spheres must also be +Q.

Let's start by considering the charge on the first sphere, Q1. We know that the charge on each sphere is directly proportional to its radius, so we can write Q1 = kR1, where k is a constant. Similarly, we can write Q2 = kR2 and Q3 = kR3. Since the total charge on all three spheres must be +Q, we can write the following equation:

Q1 + Q2 + Q3 = +Q

Substituting our expressions for Q1, Q2, and Q3, we get:

kR1 + kR2 + kR3 = +Q

Simplifying, we get:

k(R1 + R2 + R3) = Q

Solving for k, we get:

k = Q/(R1 + R2 + R3)

Now, we can substitute this value for k into our expressions for Q1, Q2, and Q3 to get:

Q1 = (Q/(R1 + R2 + R3)) * R1 = Q*R1/(R1 + R2 + R3)

Q2 = (Q/(R1 + R2 + R3)) * R2 = Q*R2/(R1 + R2 + R3)

Q3 = (Q/(R1 + R2 + R3)) * R3 = Q*R3/(R1 + R2 + R3)

Therefore, the charge on each sphere is given by:

Qi = Q*Ri/(R1 + R2 + R3)

To answer your question, yes, part (b) should use part (a) as part of the solution. This is because the potential at the center of the triangle is directly related to the charge on each sphere, which we solved for in part (a).

Part (b):
To solve for the potential at the center of the triangle