Question

If two identical conducting spheres are in contact, any excess charge will be evenly distributed between the two. Three identical metal spheres are labeled A, B, and C. Initially, A has charge q, B has charge -q/2, and C is uncharged.

a. What is the final charge on each sphere if C is touched to B, removed, and then touched to A?
b. Starting again from the initial conditions, what is the charge on each sphere if C is touched to A, removed, and then touched to B?

Answer 1

Charges on identical conducting spheres redistribute to become even when the spheres touch. Charge is conserved and shared equally among spheres upon contact. The number of electrons corresponding to a certain charge on a sphere can be calculated using the elementary charge value.

Step-by-step explanation:

When two identical conducting spheres are in contact, their charge get redistributed so that the charge is evenly divided between them. If sphere A has an initial charge of q and sphere B has an initial charge of -q/2, when sphere C, which is initially uncharged, is first touched to B, then removed, the charge on B and C will split evenly. Assuming the charge on C is now -q/4 after touching B (since no net charge is on C), when C is then touched to A, A and C will split their combined charges evenly.

For the second scenario, starting again from the initial conditions, when C is first touched to A, the charge q on A is split between A and C, making them both have a charge of q/2. Then, when C is touched to B, the charges on B and C will combine and then get split evenly, resulting in each sphere having a certain amount of charge.

To calculate the number of electrons equivalent to the charge on each sphere, we use the elementary charge of approximately 1.6 x 10^-19 C per electron.

Answer 2

a. A:
`(3)/(8)q`, B:
`-(q)/(4)`, C:
`(3)/(8)q`

When two conducting spheres touch, the charge on them is redistributed such that the potential on the two spheres is the same:

`V_1 = V_2\\(Q_1)/(C_1)=(Q_2)/(C_2)`

where Q refers to the charge and C to the capacitance of the sphere. For identical spheres, the capacitance is the same, so the previous equation becomes

`Q_1 = Q_2`

which means that the charge distributes equally on the two spheres.

Here initially we have:

Sphere A: charge of q

Sphere B: charge of -q/2

Sphere C: charge of 0

At first, sphere C is touched to sphere B. Since the total charge of the two sphere was

`-(q)/(2)+0=-(q)/(2)`

After touching each sphere will have a charge half of this value:

`q_B = q_C = (1)/(2)(-(q)/(2))=-(q)/(4)`

Then sphere C (charge of -q/4) touches sphere A (charge of +q). So the total charge is

`-(q)/(4)+q=+(3)/(4)q`

Since the charge distributes equally, each sphere will receive 1/2 of this charge:

`q_A = q_C = (1)/(2)(+(3)/(4)q)=(3)/(8)q`

So the final charge on the 3 spheres will be

A:
`(3)/(8)q`, B:
`-(q)/(4)`, C:
`(3)/(8)q`

b. A:
`(1)/(2)`, B:
`0`, C:
`0`

At first, sphere C is touched to sphere A. Since the total charge of the two sphere was

`q+0=q`

After touching each sphere will have a charge half of this value:

`q_A = q_C = (1)/(2)(q)=(q)/(2)`

Then sphere C (charge of q/2) touches sphere B (charge of -q/2). So the total charge is

`-(q)/(2)+q/2=0`

Since the charge distributes equally, each sphere will receive 1/2 of this charge, which simply means a charge of zero:

`q_B = q_C = 0`

So the final charge on the 3 spheres will be

A:
`(1)/(2)`, B:
`0`, C:
`0`