Spherical conductors connected by a wire.

[rofln]

Homework Statement

Two spherical conductors are connected by a long conducting wire, and a charge of 10.2 Micro-coulombs is placed in the combination. One sphere has a radius of 5.99 cm and the other has a radius of 7.99cm. What is the electric field near the surface of the smaller sphere. Answer in units of V/m.

Homework Equations

V = kQ/r

E = kQ/r²

k=8.99 x 10⁹

q₁ + q₂ = Q

r₁/r₂ = q₁/q₂

The Attempt at a Solution

My only issue with this problem is trying to find the individual charges on both spheres. I know that they obviously need to add up to 10.2 micro-coulombs and that the ratio of the charges should be equal to the ratio of the radii. What I've been doing is trying to find the charges using the formula for voltage for both spheres and setting them equal to each other since they are connected by the same wire. Using the net charge formula, I set q₂ to be equal to 10.2e-6 - q₁. Now after I substitute that into the voltage formula, I get:

kq₁/.0799 = (k(10.2e-6 q₁))/.0599

After isolating q₁ I get,

q₁ = 6.802838063e-6

After subtracting that from the total charge, I get the other charge to be 3.397161937e-6

Even though they add up to be 10.2e-6, I come down to the problem that the ratio does not equal that of the radii which is 1.333889816.

I've been at this for the past 2 hours trying to find mistakes in my math and I can't seem to find out what I am doing wrong. Any help is appreciated.

 

[SammyS]

Homework Statement

Two spherical conductors are connected by a long conducting wire, and a charge of 10.2 Micro-coulombs is placed in the combination. One sphere has a radius of 5.99 cm and the other has a radius of 7.99cm. What is the electric field near the surface of the smaller sphere. Answer in units of V/m.

Homework Equations

V = kQ/r

E = kQ/r²

k=8.99 x 10⁹

q₁ + q₂ = Q

r₁/r₂ = q₁/q₂

The Attempt at a Solution

My only issue with this problem is trying to find the individual charges on both spheres. I know that they obviously need to add up to 10.2 micro-coulombs and that the ratio of the charges should be equal to the ratio of the radii. What I've been doing is trying to find the charges using the formula for voltage for both spheres and setting them equal to each other since they are connected by the same wire. Using the net charge formula, I set q₂ to be equal to 10.2e-6 - q₁. Now after I substitute that into the voltage formula, I get:

kq₁/.0799 = (k(10.2e-6 q₁))/.0599

After isolating q₁ I get,

q₁ = 6.802838063e-6

After subtracting that from the total charge, I get the other charge to be 3.397161937e-6

Even though they add up to be 10.2e-6, I come down to the problem that the ratio does not equal that of the radii which is 1.333889816.

I've been at this for the past 2 hours trying to find mistakes in my math and I can't seem to find out what I am doing wrong. Any help is appreciated.

Hello rofln. Welcome to PF !

The two spheres are connected by a (conducting) wire. So in essence, all three objects for one conductor. What's true of the potential throughout a conductor under eltro-static conditions?

 

[rofln]

The potential everywhere is the same and constant everywhere inside of it.

 

[SammyS]

The potential everywhere is the same and constant everywhere inside of it.

So the potential just outside the surface of the larger sphere is equal to the potential just outside the surface of the smaller sphere. Correct?

So, what's the ratio of the charge on the spheres?

 

[rofln]

The ratio of the charges should be the same as the potential?

 

[SammyS]

The potentials are the same. The radii are different.