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Galina-37 [17]
4 years ago
15

A parallel-plate capacitor is formed from two 7.2 cm-diameter electrodes spaced 1.8 mm apart. The electric field strength inside

the capacitor is 3.0×106N/C. What is the magnitude of the charge (in nC) on each electrode?
Physics
1 answer:
MatroZZZ [7]4 years ago
6 0

Answer: 432 nC

Explanation: In order to calculate the charge,  we know that the electric field corresponding for two parallel-plate is given by:

E=Q/(area*e0)

In our case E=3.0x10^6 N/C

Area = Pix (0,072 m)^2

e0 vacuum permittivity 8,85x 10^-12

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Read 2 more answers
Pls help ASAP
12345 [234]

Answer:

a. 4.733 × 10⁻¹⁹ J = 2.954 eV b i. yes ii. 0.054 eV = 8.651 × 10⁻²¹ J

Explanation:

a. Find the energy of the incident photon.

The energy of the incident photon E = hc/λ where h = Planck's constant = 6.626 × 10⁻³⁴ Js, c = speed of light = 3 × 10⁸ m/s and λ = wavelength of light = 420 nm = 420 × 10⁻⁹ m

Substituting the values of the variables into the equation, we have

E = hc/λ

= 6.626 × 10⁻³⁴ Js × 3 × 10⁸ m/s ÷ 420 × 10⁻⁹ m

= 19.878 × 10⁻²⁶ Jm  ÷ 420 × 10⁻⁹ m

= 0.04733 × 10⁻¹⁷ J

= 4.733 × 10⁻¹⁹ J

Since 1 eV = 1.602 × 10⁻¹⁹ J,

4.733 × 10⁻¹⁹ J = 4.733 × 10⁻¹⁹ J × 1 eV/1.602 × 10⁻¹⁹ J = 2.954 eV

b. i. Is this energy enough for an electron to leave the atom

Since E = 2.954 eV is greater than the work function Ф = 2.9 eV, an electron would leave the atom. So, the answer is yes.

ii. What is its  maximum energy?

The maximum energy E' = E - Ф = 2.954 - 2.9

= 0.054 eV

= 0.054 × 1 eV

= 0.054 × 1.602 × 10⁻¹⁹ J

= 0.08651  × 10⁻¹⁹ J

= 8.651 × 10⁻²¹ J

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