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Flura [38]
3 years ago
8

g Where is the dielectric constant and A is the area of the capacitor plates. The charge q and voltage Vc across the capacitor p

lates are related by The electric field produces a force Fe on the movable plate (b) that acts to reduce the plate gap Derive non-linear state equations that describe the electromechanical dynamics of this system.

Physics
1 answer:
aev [14]3 years ago
6 0

Answer:

<u>Capacitor and the dielectrics:</u>

"The capacitor is a charge,q storing device as it is subjected to the potential difference between the different plates and thus results in the difference of charge decomposition between the two plates."

<u>Formula:</u>C= ∈°(A/d),

<u>Unit:</u>  Farad,F.

  • <u>Purpose of the Dielectrics:</u>

In order to store or hold the optimum number of charge particles,q between the plates we have a number of insulators which are used as the medium between the two plates.While, it separates both the plates from one another.

Explanation:

<u>Dielectric Constants in the capacitors:</u>

In order to hold the number of charges in between the plates of the capacitor a specific medium or substrate is placed(mostly the insulators). As it covers the area between the two capacitors, while it can decrease and increase by the action of the movable plate of the capacitor.

Now, it increase the capcitative power of the capacitor by decreasing the potential difference between the two plates, along with which the electromagnetic field is also greatly reduced.As the interaction between the two plates is greatly reduced, as they are covered and well protected by the sheet or membrane of insulting medium around them.

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Answer:

805.48N/m

Explanation:

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3 0
2 years ago
An electric motor rotating a workshop grinding wheel at 1.06 102 rev/min is switched off. Assume the wheel has a constant negati
kvasek [131]

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3 years ago
A boat moves with a speed of +2.5 m/s in a direction 25° north of east. If the mass of the boat is 15,000 kg, what is the moment
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Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc
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Given

car A had a head start of D_A

and it starts at x=0 and t=0

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4 years ago
In an experiment, a variable, position-dependent force F(x)F(x) is exerted on a block of mass 1.0kg1.0kg that is moving on a hor
leonid [27]

Answer:

The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(C) is correct option.

Explanation:

Given that,

Mass of block = 1.0 kg

Dependent force = F(x)

Frictional force = F(f)

Suppose, the following information would students need to test the hypothesis,

(A) The function F(x) for 0 < x < 5 and the value of F(f).

(B) The function a(t) for the time interval of travel and the value of F(f).

(C) The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(D) The function a(t) for the time interval of travel, the time it takes the block to move 5 m, and the value of F(f).

(E) The block's initial velocity, the time it takes the block to move 5 m, and the value of F(f).

We know that,

The work done by a force is given by,

W=\int_{x_{0}}^{x_{f}}{F(x)\ dx}.....(I)

Where, F(x) = net force

We know, the net force is the sum of forces.

So, \sum{F}=ma

According to question,

We have two forces F(x) and F(f)

So, the sum of these forces are

F(x)+(-F(f))=ma

Here, frictional force is negative because F(f) acts against the F(x)

Now put the value in equation (I)

W=\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}

We need to find the value of \int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}

Using newton's second law

\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{ma\ dx}...(II)

We know that,

Acceleration is rate of change of velocity.

a=\dfrac{dv}{dt}

Put the value of a in equation (II)

\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{m\dfrac{dv}{dt}dx}

\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{v_{0}}^{v_{f}}{mv\ dv}

\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}

Now, the work done by the net force on the block is,

W=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}

The work done by the net force on the block is equal to the change in kinetic energy of the block.

Hence, The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(C) is correct option.

7 0
3 years ago
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