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Ipatiy [6.2K]
3 years ago
11

A 1 uF parallel-plate capacitor is charged to 12 V and then disconnected from the voltage supply. The plate separation distance

is doubled and the dielectric is replaced by a material with another that has one quarter of the former relative permittivity. Find the final charge on the capacitor's positive plate and the final voltage difference between its terminals. 2. A parallel-plate capacitor consists of two equally thick layers of dielectric sandwiched between 1 cm X 2 cm rectangular plates that are separated by 200 um. One layer of dielectric has permittivity of 5 uF/m. The other has twice that value. Find the energy stored in the capacitor when its leads are connected to a 60 V supply.
Physics
1 answer:
nlexa [21]3 years ago
6 0

Answer:

22

Explanation:

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Makovka662 [10]

Answer:

1) n = 4.47*10^12 photons

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Explanation:

This is a problem about the photoelectric effect.

1) In order to calculate the number of photons that impact the metal, you take into account the power of the light, which is given by:

P=\frac{E}{t}=1.4*10^{-6}\frac{J}{s}     (1)

Furthermore you calculate the energy of a photon with a wavelength of 600nm, by using the following formula:

E_p=h\frac{c}{\lambda}         (2)

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You replace the values of the parameters in the equation (2):

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The number of photons is 4.47*10^12

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Ep: energy of the photons

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You first convert the energy of the photons to eV:

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You replace in the equation (3):

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The kinetic energy of the electrons emitted by the metal is 0.25 eV

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