Two Wired Up Metallic Spheres


by jaguar7
Tags: metallic, spheres, wired
jaguar7
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#1
Feb19-11, 10:15 PM
P: 42
1. The problem statement, all variables and given/known data

In the figure, r1 = 2 and r2 = 8 cm. Before the conducting spheres are connected by the wire a charge of 6.010-7 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)

2. Relevant equations



3. 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
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gneill
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#2
Feb19-11, 10:33 PM
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What is the electric potential just above the surface of a charged conducting sphere?
jaguar7
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#3
Feb19-11, 10:57 PM
P: 42
Quote Quote by gneill View Post
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.

Piyu
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#4
Feb19-11, 11:01 PM
P: 45

Two Wired Up Metallic Spheres


Is potential or electric field the one that decides force per unit charge?
gneill
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#5
Feb19-11, 11:14 PM
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A formula would be nice. Something involving the charge on the sphere, q, and the radius of the sphere, R.
jaguar7
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#6
Feb19-11, 11:17 PM
P: 42
Quote Quote by Piyu View Post
Is potential or electric field the one that decides force per unit charge?
It's electric field.
jaguar7
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#7
Feb19-11, 11:23 PM
P: 42
Quote Quote by gneill View Post
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...
jaguar7
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#8
Feb19-11, 11:30 PM
P: 42
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 +
nestharus
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#9
Feb22-12, 12:21 AM
P: 1
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.


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