Two Wired Up Metallic Spheres

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In summary, the problem involves calculating the charge on a smaller sphere after it is connected to a larger sphere via a wire. The charge on the smaller sphere is equal to the fraction of the surface area of the smaller sphere over the total surface area of both spheres multiplied by the total charge. The key to solving this problem is knowing that the electric potential for both spheres will be the same due to the wire connecting them. The formula for electric potential symmetry for spherical shapes can be used to find the charge on the smaller sphere.
  • #1
jaguar7
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Homework Statement



In the figure, r1 = 2 and r2 = 8 cm. Before the conducting spheres are connected by the wire a charge of 6.0×10-7 C is placed on the smaller sphere while the larger sphere is uncharged. Calculate the charge on the smaller sphere after the wire is connected. Assume that the separation of the spheres is very large compared to their radii.

(Figure shows a small sphere on the left with a wire attaching it to a larger sphere on the right)

Homework Equations


The Attempt at a Solution

When the spheres are wired together the charge travels to the larger one also, and is now spread evenly along the surface of both spheres. The charge on the smaller one should be the fraction of the surface area of the small sphere over the total surface area of both spheres times the total charge, q.

q * 4pir1^2 / (4pir1^2 + 4pir2^2) = qr1^2 / (r1^2 + r2^2).

--Instructor reply: "The charge distribution is not a simple function of the ratio of the surface areas. The key for this problem is that the wire means that the skins for both spheres will be at the same electric potential."

--- me, now: I don't know what to do... T_T
 
Last edited:
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  • #2
What is the electric potential just above the surface of a charged conducting sphere?
 
  • #3
gneill said:
What is the electric potential just above the surface of a charged conducting sphere?

I think it's pretty high because there is a lot of force acting on a test charge at that point, and so there is a lot of potential energy per charge, which is electric potential.
 
  • #4
Is potential or electric field the one that decides force per unit charge?
 
  • #5
A formula would be nice. Something involving the charge on the sphere, q, and the radius of the sphere, R.
 
  • #6
Piyu said:
Is potential or electric field the one that decides force per unit charge?

It's electric field.
 
  • #7
gneill said:
A formula would be nice. Something involving the charge on the sphere, q, and the radius of the sphere, R.

yeah, I know. Do we have anything that has area and q? I don't know. We have these formulas for electric potential symmetry --for shereical: delV = kQ (1/rf - 1/ri)

and planar, but I don't think it's planar. I don't know what equation to use..

I can't find any equations in my book relating V to A and Q...

or E to A and Q either, because E = kq/r^2 for outside the surface of a conducting sphere...
 
  • #8
My instructor says that the electric potential for both spheres is the same...

this ended in the correct answer... typed it out because I got it wrong at first because I made an algebra mistake, caught while typing it...

so V1 = V2
kq1/r1 = kq2/r2, q1 + q2 = qT --> q2 = qT - q1

q1 = (qT - q1)r1/r2

q1 = r1qT/r2 - r1q1/r2

q1 +
 
Last edited:
  • #9
Ok, just wanted to add some feed back for people who didn't get what you did ; )

First, think of qTotal as the sum of q1 old + q2 old. Remember that q2 old is 0, so qTotal is just q1 old.

We now look at our equation
dv1 = dv2
k*q1/r1 = k*q2/r2

We are trying to solve for q1. qTotal is still equal to q1 new + q2 new. This means that we can substitute it in.

q2 = (r2/r1)*q1

Remember qTotal = q1+q2

q2 = qTotal - q1

qTotal - q1 = (r2/r1)*q1
qTotal = q1 + (r2/r1)*q1
qTotal = q1 (1 + r2/r1)

q1 = qTotal/(1 + r2/r1)

You know qTotal, r2, and r1, so just plug and chug.
 

1. What are "Two Wired Up Metallic Spheres"?

"Two Wired Up Metallic Spheres" refer to a scientific experiment in which two metallic spheres are connected by a wire and charged with electricity.

2. What is the purpose of this experiment?

The purpose of this experiment is to demonstrate the principles of electric charge and to observe the behavior of charged objects.

3. How does this experiment work?

In this experiment, the two metallic spheres are first connected by a wire, creating a closed circuit. One sphere is then given a positive charge, while the other is given a negative charge. This creates an electric field between the spheres, causing them to repel each other.

4. What can be learned from this experiment?

This experiment can teach us about electric charge, electric fields, and the behavior of charged objects. It can also demonstrate the concept of equilibrium, as the spheres will eventually stop moving when the charges are equal.

5. Are there any real-life applications for this experiment?

Yes, there are several real-life applications for this experiment. Understanding electric charge and electric fields is crucial in fields such as electronics, energy production, and medical technology. This experiment can also be used to demonstrate the concept of static electricity, which has applications in industries such as printing and painting.

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