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tensa zangetsu [6.8K]
2 years ago
13

Two identical air-filled parallel-plate capacitors C1 and C2 are connected in series to a battery that has voltage V. The charge

on each capacitor is Q0. While the two capacitors remain connected to the battery, a dielectric with dielectric constant K>1 is inserted between the plates of capacitor C1, completely filling the space between them.
Physics
1 answer:
Llana [10]2 years ago
5 0

Complete question:

Two identical air-filled parallel-plate capacitors C₁ and C₂ are connected in series to a battery that has voltage V. The charge on each capacitor is Qo. While the two capacitors remain connected to the battery, a dielectric with dielectric constant K > 1 is inserted between the plates of capacitor C1, completely filling the space between them. What is the charge on capacitor C1 after the dielectric is inserted? Express your answer in terms of K and Qo.

Answer:

The charge on capacitor after the dielectric is inserted is Qo( k + 1)

Explanation:

Given;

two identical capacitor, C₁ and C₂

charge on each capacitor, Q = Qo

Voltage of the battery, = V

dielectric constant K > 1

After inserting dielectric constant K, the equivalent capacitance in parallel becomes;

C.equ. = KC₁ + C₂

Charge, on capacitor Qo = C.equ.V

Q = V(KC₁ + C₂)

Q = KC₁V + C₂V

since, the charge on each capacitor is Qo, then

Q = KQo + Qo

Q = Qo( k + 1)

Therefore, the charge on capacitor after the dielectric is inserted is Qo( k + 1), in terms of K and Qo.

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Extraneous

Explanation:

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2 years ago
Three charges are enclosed inside a spherical closed surface. The net flux through the surface is −216 N · m2/C. If two of the c
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Answer:

q₃ = -4.81 nC

Explanation:

We can use the Gauss Law here:

∅ = q/∈₀

where,

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q = total charge enclosed inside sphere = ?

∈₀ = permittivity of free space = 8.85 x 10⁻¹² C/N.m²

Therefore,

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q = (-216 N.m²/C)(8.85 x 10⁻¹² C²/N.m²)

q = - 1.91 nC

So, the total charge will be sum of all three charges:

q = q₁ + q₂ + q₃

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q₃ = - 1.91 nC - 1.74 nC - 1.16 nC

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Explanation:

Formula to determine the critical crack is as follows.

          K_{IC} = \gamma \sigma_{f} \sqrt{\pi \times a}

  \gamma = 1,     K_{IC} = 24.1

  [/tex]\sigma_{y}[/tex] = 570

and,   \sigma_{f} = 570 \times \frac{3}{4}

                       = 427.5

Hence, we will calculate the critical crack length as follows.

      a = \frac{1}{\pi} \times (\frac{K_{IC}}{\sigma_{f}})^{2}

        = \frac{1}{3.14} \times (\frac{24.1}{427.5})^{2}

       = 10.13 \times 10^{-4}

Therefore, largest size is as follows.

            Largest size = 2a

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Where:

k is the elastic constant of the spring. The higher its value, the more work it will cost to stretch the spring.

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According to this, we have a spring where only the force due gravity is applied.

In other words, the force applied is the weigth W of the block:

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