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scoray [572]
3 years ago
13

A capacitor with plates separated by distance d is charged to a potential difference ΔVC. All wires and batteries are disconnect

ed, then the two plates are pulled apart (with insulated handles) to a new separation of distance 2d. Part A Does the capacitor charge Q change as the separation increases? If so, by what factor? If not, why not?

Physics
2 answers:
nevsk [136]3 years ago
8 0

Answer:

When the separation between the capacitor plates is increased to 2d, and if the battery is disconnected, the charge between the condenser plates does not change, it is the same, and this is represented by the following expression: Q'= Q

Explanation:

fgiga [73]3 years ago
5 0

Answer:

Yes, the capacitor's Q load varies inversely proportional to the distance between plates.

Explanation:

In the attached files you see the inverse relationship between capacity and distance between plates "d".

In the following formula we see its relationship with the "Q" load

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A particle with charge 6 mC moving in a region where only electric forces act on it has a kinetic energy of 1.9000000000000001 J
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Answer:

16.9000000000000001 J

Explanation:

From the given information:

Let the initial kinetic energy from point A be K_A = 1.9000000000000001 J

and the final kinetic energy from point B be K_B = ???

The charge particle Q = 6 mC = 6 × 10⁻³ C

The change in the electric potential from point B to A;

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According to the work-energy theorem:

-Q × ΔV = ΔK

-Q \times ( V_B - V_A) = (K_B - K_A)

-(6\times 10^{-3}\ C) \times ( -2.5 \times 10^3) = (K_B - 1.9000000000000001 \ J)

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K_B = 15+ 1.9000000000000001 \ J

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3 years ago
One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction
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Answer:

F = 156.3 N

Explanation:

Let's start with the top block, apply Newton's second law

         F - fr = 0

         F = fr

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Now we can work  with the bottom block

In this case we have two friction forces, one between the two blocks and the other between the block and the surface. In the exercise, indicate that the two friction coefficients are equal

we apply Newton's second law

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as the two blocks are identical

        N = 2W

X axis

        F - fr₁ - fr₂ = 0

        F = fr₁ + fr₂

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a

s the normal in the lower block of twice the friction force is

          fr₂ = μ 2N

          fr₂ = 2 μ N

          fr₂ = 2 fr₁

we substitute

          F = fr₁ + 2 fr₁

          F = 3 fr₁

          F = 3  52.1

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