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DerKrebs [107]
2 years ago
3

A conducting sphere of radius R carries positive charge q. Calculate the amount of work that would be required to move a small p

ositive test charge q0 slowly from r=5R to r=3R. Assume that the presence of q0 has no effect on how the charge q is distributed over the sphere.
Express your answer in terms of the electric constant ϵ0 and some or all of the variables q, q0, and R.
Physics
1 answer:
kicyunya [14]2 years ago
6 0

Answer:

Explanation:

Given

Radius of Conducting Sphere is R with Positive charge q

initially small sphere is at a distance of r=5 R and slowly move to r=3 R

Electric Potential Energy Between two Charged Particle

U=k\cdot \frac{q_1q_2}{r^2}

Initial Potential Energy

U_1=k\cdot \frac{qq_0}{(5R)^2}

U_1=k\cdot \frac{qq_0}{25R^2}

at r=3R

U_2=k\cdot \frac{qq_0}{(3R)^2}

U_2=k\cdot \frac{qq_0}{9R^2}

Work Done =U_2-U_1

W=k\cdot \frac{qq_0}{9R^2}-k\cdot \frac{qq_0}{25R^2}

W=k\cdot \frac{qq_0}{R^2}\left [ \frac{1}{9}-\frac{1}{25}\right ]

W=k\cdot \frac{qq_0}{R^2}\cdot \left [ \frac{16}{225}\right ]

W=\frac{1}{4\pi \epsilon _0}\cdot \frac{qq_0}{R^2}\cdot \left [ \frac{16}{225}\right ]

W=\frac{4qq_0}{225\pi \epsilon _0R^2}        

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Ancient Greek philosophers spent lots of time thinking about science and imaging explanations for the natural world. What part o
Illusion [34]

Answer:

Testing

Explanation:

Ancient Greek philosophers lived with the ideology to simply contemplate life. This means that their whole life revolved around thinking and questioning everything. This would include creative thinking, because they would sometimes come up with theories which require creativeness. They would often debate with their friends as to why their theory should be accepted or what their opinions were on the matter. More often than not, they argued a lot, and many philosophers went against some powerful people in the community and some were even sentenced to death.

The main process they didn't/couldn't do was the testing. They could never test certain theories because they did not have the means to.

4 0
2 years ago
The different in size of each of the rope's pullers, correspond to a difference in the magnitude of the applied force, such that
olga55 [171]

Answer:

F = - 50 N

Hence, the magnitude of resultant force is 50 N and its direction is leftwards.

Explanation:

The magnitude of the resultant force is always equal to the sum of all forces. While, the direction of resultant force will be equal to the direction of the force with greater magnitude:

Resultant\ Force = F = F_{1} - F_{2}

considering right direction to be positive:

F₁ = Force applied on right rope = 150 N

F₂ = Force applied on left rope = 200 N

Therefore, the resultant force can be found by using these values in equation:

F = 150\ N - 200\ N

<u>F = - 50 N</u>

<u>Hence, the magnitude of resultant force is 50 N and its direction is leftwards.</u>

5 0
2 years ago
A charge of 87.6 pC is uniformly distributed on the surface of a thin sheet of insulating material that has a total area of 65.2
LiRa [457]

Answer:

60.8 cm²

Explanation:

The charge density, σ on the surface is σ = Q/A where q = charge = 87.6 pC = 87.6 × 10⁻¹² C and A = area = 65.2 cm² = 65.2 × 10⁻⁴ m².

σ = Q/A = 87.6 × 10⁻¹² C/65.2 × 10⁻⁴ m² = 1.34 × 10⁻⁸ C/m²

Now, the charge through the Gaussian surface is q = σA' where A' is the charge in the Gaussian surface.

Since the flux, Ф = 9.20 Nm²/C and Ф = q/ε₀ for a closed Gaussian surface

So, q = ε₀Ф = σA'

ε₀Ф = σA'

making A' the area of the Gaussian surface the subject of the formula, we have

A' = ε₀Ф/σ

A' = 8.854 × 10⁻¹² F/m × 9.20 Nm²/C ÷ 1.34 × 10⁻⁸ C/m²

A' = 81.4568/1.34 × 10⁻⁴ m²

A' = 60.79 × 10⁻⁴ m²

A' ≅ 60.8 cm²

6 0
2 years ago
A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate
diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

\Delta S = \frac{Q}{T}

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

W = Q_{source}-Q_{sink}

According to the data given we have to,

Q_{source} = 200000Btu

T_{source} = 1500R

Q_{sink} = 100000Btu

T_{sink} = 600R

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}

\Delta S_{sink} = \frac{100000}{600}

\Delta S_{sink} = 166.67Btu/R

On the other hand,

\Delta S_{source} = \frac{Q_{source}}{T_{source}}

\Delta S_{source} = \frac{-200000}{1500}

\Delta S_{source} = -133.33Btu/R

The total change of entropy would be,

S = \Delta S_{source}+\Delta S_{sink}

S = -133.33+166.67

S = 33.34Btu/R

Since S\neq   0 the heat engine is not reversible.

PART B)

Work done by heat engine is given by

W=Q_{source}-Q_{sink}

W = 200000-100000

W = 100000 Btu

Therefore the work in the system is 100000Btu

4 0
2 years ago
How is earth outer layer different from a cracked hard-boiled egg?
Aleksandr [31]

Answer:

The Earth holds livings things and even the layer has more complex layers in it (Also it's not white) And of course there are no oceans.

7 0
2 years ago
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