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qwelly [4]
3 years ago
8

If two identical conducting spheres are in contact, any excess charge will be evenly distributed between the two. Three identica

l 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?
Physics
1 answer:
Ipatiy [6.2K]3 years ago
3 0

a. A: \frac{3}{8}q, B: -\frac{q}{4}, C: \frac{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\\\frac{Q_1}{C_1}=\frac{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

-\frac{q}{2}+0=-\frac{q}{2}

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

q_B = q_C = \frac{1}{2}(-\frac{q}{2})=-\frac{q}{4}

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

-\frac{q}{4}+q=+\frac{3}{4}q

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

q_A = q_C = \frac{1}{2}(+\frac{3}{4}q)=\frac{3}{8}q

So the final charge on the 3 spheres will be

A: \frac{3}{8}q, B: -\frac{q}{4}, C: \frac{3}{8}q

b. A: \frac{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 = \frac{1}{2}(q)=\frac{q}{2}

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

-\frac{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: \frac{1}{2}, B: 0, C: 0

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