Physics, 30.01.2020 08:43 cylertroutner

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?

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Answer from: kimboblack4192

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: , C:

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: , C:

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Answer from: Quest

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