Physics : asked on kamila20394
 03.10.2022

A parallel-plate air capacitor is made from two plates 0.200{\rm m} square, spaced 0.800 {\rm cm} apart. It is connected to a 120 {\rm V} battery. If the plates are pulled apart to aseparation of 1.60 {\rm cm}, suppose the battery remains connected while theplates are pulled apart. What is the capacitance?

C =2.21×10−11{\rm F}

Part B

What is the charge on each plate?

Enter your answer as two numbers,separated with a comma.

Part C

What is the electric field between theplates?

E =7500 {\rm V/m}

Part D

What is the energy stored in thecapacitor?

U =1.59×10−7{\rm J}

. 1

Step-by-step answer

24.06.2023, solved by verified expert
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A)  C = 4.425 10⁻¹³ F,  B) Q = 6.31 10⁻¹¹ C,  C)  E = 8.9 10⁴ N / C,  

D)  u_{E} = 3.19 10⁻⁹ J

Explanation:

Part A. Capacitance is

           C = ε₀ A / d

in this case distance is d = 0.800 cm = 0.800 10⁻² m and the area of the plates is x = 0.200 cm= 0.2 10⁻² m, all the quantities must be in the SI system for the result to be in Farads

           A = x²

           A = (0.2 10⁻²)²

           A = 4 10⁻⁶ m

let's calculate

          C = 8.85 10⁻¹² 4 10⁻⁶ / 0.8 10⁻²

          C = 44.25 10⁻¹⁴ F

          C = 4.425 10⁻¹³ F

Part B. The charge on each plate is

          Q = C ΔV

           Q = 4.425 10⁻¹³ 120

           Q = 6.31 10⁻¹¹ C

Part C. the electric field of a plate is

           E = σ / 2ε₀

where the charge density is

            σ = Q / A

we substitute

             E =A parallel-plate air capacitor is made from two, №17887357, 03.10.2022 12:27

       

let's calculate

             E = A parallel-plate air capacitor is made from two, №17887357, 03.10.2022 12:27

             E = 8.9 10⁴ N / C

Part D. stored energy

              A parallel-plate air capacitor is made from two, №17887357, 03.10.2022 12:27 = ½ C V²

               u_{E} = ½ 4.425 10⁻¹³ 120²

               u_{E} = 3.19 10⁻⁹ J

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Faq

Physics
Step-by-step answer
P Answered by Master

Answer:

see below.

Step-by-step explanation:

To solve this problem, we can use the conservation of energy and conservation of momentum principles.

Conservation of energy:

The total initial energy is the rest energy of the proton and neutron, which is given by:

Ei = (mp + mn)c^2

where mp and mn are the masses of the proton and neutron, respectively, and c is the speed of light.

The total final energy is the rest energy of the deuteron plus the energy of the gamma ray, which is given by:

Ef = (md)c^2 + Eg

where md is the mass of the deuteron and Eg is the energy of the gamma ray.

According to the conservation of energy principle, the initial energy and final energy must be equal, so we have:

Ei = Ef

(mp + mn)c^2 = (md)c^2 + Eg

Conservation of momentum:

The total initial momentum is zero because the proton and neutron are at rest. The total final momentum is the momentum of the deuteron and the momentum of the gamma ray. Since the gamma ray is massless, its momentum is given by:

pg = Eg/c

where pg is the momentum of the gamma ray.

According to the conservation of momentum principle, the total final momentum must be equal to zero, so we have:

0 = pd + pg

where pd is the momentum of the deuteron.

Solving for md and pd:

From the conservation of energy equation, we can solve for md:

md = (mp + mn - Eg/c^2)/c^2

Substituting this expression into the conservation of momentum equation, we get:

pd = -pg = -Eg/c

Substituting the given values, we have:

mp = 1.6726 × 10^-27 kg mn = 1.6749 × 10^-27 kg Eg = 2.2 × 10^6 eV = 3.52 × 10^-13 J

Using c = 2.998 × 10^8 m/s, we get:

md = (1.6726 × 10^-27 kg + 1.6749 × 10^-27 kg - 3.52 × 10^-13 J/(2.998 × 10^8 m/s)^2)/(2.998 × 10^8 m/s)^2 = 3.3435 × 10^-27 kg

pd = -Eg/c = -(3.52 × 10^-13 J)/(2.998 × 10^8 m/s) = -1.1723 × 10^-21 kg·m/s

Therefore, the mass of the deuteron is 3.3435 × 10^-27 kg, and its momentum is -1.1723 × 10^-21 kg·m/s.

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