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2 answers:
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Answer:
Explanation:
The capacitance of a parallel-plate capacitor is given by
where is the vacuum permittivity, A is the surface area of the plates, d their separation.
We also know the following relationship
where Q is the charge stored on the capacitor and V the potential difference between the plates.
Combining the two equations,
We also know that for a uniform electric field (such as the one between the plates of a parallel-plate capacitor), we have
where E is the magnitude of the electric field. Substituting into the previous equation and re-arranging it,
For the capacitor in the problem:
is the area of the plates
is the maximum electric field before a spark is produced
Solving for Q, we find the maximum charge that can be added before that occurs:
Answer:
first minimum distance from the center = 0.417 m m
Explanation:
given,
wavelength (λ) = 506 nm
d = 0.075 mm = 0.000075 m
tanθ = x /2
d sinθ = m λ
m=1 ( 1 st minimum )
0.000075 × sin θ = 506 × 10⁻⁹
sin θ =
sin θ = 0.00675
θ = sin⁻¹(0.00675)
θ = 0.00675
x = 0.000417 m
hence, the first minimum distance from the center = 0.417 m m