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navik [9.2K]
3 years ago
14

The paper dielectric in a paper-and-foil capacitor is 0.0800 mm thick. Its dielectric constant is 2.50, and its dielectric stren

gth is 50.0 MV/m. Assume that the geometry is that of a parallel-plate capacitor, with the metal foil serving as the plates. (a) What area of each plate is required for a 0.200 wF capacitor? (b) If the electric field in the paper is not to exceed one-half the dielectric strength, what is the maximum potential difference that can be applied across the capacitor?
Physics
1 answer:
jeka943 years ago
8 0

Answer:

a) 0.723 m²

b) 2000V

Explanation:

Given that

Thickness of the capacitor, d = 0.08*10^-3 m

Dielectric constant of the capacitor, k = 2.5

Dielectric strength of the capacitor, E = 50*10^6

Capacitance of the capacitor, C = 0.2*10^-6

Permittivity of free space, E• = 8.85*10^-12

a)

The area, A is given by the formula

A = (C * d) / (k * E•)

A = (0.2*10^-6 * 0.08*10^-3) / (2.5 * 8.85*10^-12)

A = 1.6*10^-11 / 2.213*10^-11

A = 0.723 m²

b)

Potencial difference, V is given by the formula

V = E * d

V = 1/2 * 50*10^6 * 0.08*10^-3

V = 1/2 * 4000

V = 2000 V

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Answer:

39.40 MeV

Explanation:

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when we apply this values using the relativistic energy-momentum relation

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Also in a nuclear confinement ( E, P >> mc )

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K = h^2 / 2ma^2 = 1.52 GeV

7 0
3 years ago
We want to construct a solenoid with a resistance of 4.30 Ω and generate a magnetic field of 3.70 × 10−2 T at its center when ap
marshall27 [118]

Answer with Explanation:

We are given that

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Radius of wire,r=\frac{d}{2}=\frac{0.5\times 10^{-3}}{2}=0.25\times 10^{-3} m

1mm=10^{-3} m

Radius of solenoid,r'=1 cm=1\times 10^{-2} m

1 cm=10^{-2} m

Resistivity of copper,\rho=1.68\times 10^{-8}\Omega m

We know that

R=\frac{\rho l}{A}

Where A=\pi r^2

Using the formula

4.3=\frac{1.68\times 10^{-8}\times l}{\pi(0.25\times 10^{-3})^2}

l=\frac{4.3\times \pi(0.25\times 10^{-3})^2}{1.68\times 10^{-8}}=50.23 m

Number of turns of wire=\frac{l}{2\pi r'}

Number of turns of wire=\frac{50.26}{2\pi(1\times 10^{-2}}=800

Hence, the number of turns of the  solenoid,N=799

Magnetic field in solenoid,B=\mu_0 nI

3.7\times 10^{-2}=4\pi\times 10^{-7} n\times 4.6

n=\frac{3.7\times 10^{-2}}{4\times 3.14\times 10^{-7}\times 4.6}

n=6404 turns/m

n=\frac{N}{L}

L=\frac{N}{n}

L=\frac{799}{6404}

L=0.125 m=0.125\times 100=12.5 cm

Length of solenoid=12.5 cm

1m=100 cm

8 0
3 years ago
What is the magnitude of the change in potential energy of the block-spring system when it travels from its lowest vertical posi
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the potential energy at the highest point is

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the potential energy at the lowest point is

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instead in this energy it is

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4 0
3 years ago
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lisabon 2012 [21]

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Hope that helped! :)

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3 years ago
A plastic film moves over two drums. During a 4-s interval the speed of the tape is increased uniformly from v0 = 2ft/s to v1 =
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Answer:

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v = \omega R\\a = \alpha R

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Question 2)

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5 0
3 years ago
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