1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
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

You might be interested in
A 1.0 kg ball falls from rest a distance of 19.6 m.
Nimfa-mama [501]

Answer:

192.08J

19.6m/s

Explanation:

Since there will be no potential energy when the ball is on the ground, the change in potential energy is equal to the potential energy at the start when the ball is 19.6m above the ground.

PE=mgh

=(1)(9.8)(19.6)

=192.08J

v²=u²+2as, where v is the final velocity, u is initial velocity, a is acceleration and s is distance. Initial velocity is 0 since it starts at rest.

v²=u²+2as

v²=0²+2(9.8)(19.6)

v=√384.16

=19.6m/s

6 0
3 years ago
A 0.153 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.700 m/s. It has a head-on col
DedPeter [7]

Answer:

3.1216 m/s.

Explanation:

Given:

M1 = 0.153 kg

v1 = 0.7 m/s

M2 = 0.308 kg

v2 = -2.16 m/s

M1v1 + M2v2 = M1V1 + M2V2

0.153 × 0.7 + 0.308 × -2.16 = 0.153 × V1 + 0.308 × V2

= 0.1071 - 0.66528 = 0.153 × V1 + 0.308 × V2

0.153V1 + 0.308V2 = -0.55818. i

For the velocities,

v1 - v2 = -(V1 - V2)

0.7 - (-2.16) = -(V1 - V2)

-(V1 - V2) = 2.86

V2 - V1 = 2.86. ii

Solving equation i and ii simultaneously,

V1 = 3.1216 m/s

V2 = 0.2616 m/s

8 0
3 years ago
The moon Phobos orbits Mars
podryga [215]

Explanation:

For a circular orbit v= \sqrt{\frac{G.m}{r} } with G = 6.6742 × 10^{-11}

Given m = 6.42 x 10^23 kg and  r=9.38 x 10^6 m

=> v = 2137.3 m/s

I hope this is the correct way to solve

3 0
3 years ago
When an object in simple harmonic motion is at its maximum displacement, its____________ is also at a maximum.
Ivan

When an object in simple harmonic motion is at its maximum displacement, its <u>acceleration</u> is also at a maximum.

<u><em>Reason</em></u><em>: The speed is zero when the simple harmonic motion is at its maximum displacement, however, the acceleration is the rate of change of velocity. The velocity reverses the direction at that point therefore its rate of change is maximum at that moment. thus the acceleration is at its maximum at this point</em>

<em />

Hope that helps!

7 0
3 years ago
What if you were a mother for a day boys and girls
Dominik [7]
I’m am a mother no drama !!!!

Boom
6 0
3 years ago
Other questions:
  • A spider hangs by a strand of silk at an eye level 20 cm in front of a plane mirror. You are behind the spider, 65 cm from the m
    11·1 answer
  • Define fission and give an example of a reaction
    7·2 answers
  • In which one or more of the following is the earth assumed to be the center of the universe?
    15·2 answers
  • Which of the following is not a part of a wave?
    15·2 answers
  • Julian is late to class and sees a list on the board: -are explanations of observations and hypotheses -are based on many experi
    8·2 answers
  • What will happen if a low massive main sequence star runs out of hydrogen fuel?​
    10·1 answer
  • James went to the doctor for a check-up and found his weight to be 1,000 N. How much mass does James have?
    7·1 answer
  • What type of metals are known as ferromagnetic metals
    12·1 answer
  • Convert 75.0 degrees Celsius into Fahrenheit
    12·1 answer
  • The volume in the pump when the pump piston is all the way down represents the _______
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!