A parallel-plate air capacitor with a capacitance of 260 pF has a charge of magnitude 0.155 μC on each plate. The plates have a
1 answer:
Answer:
The potential difference between the plates is 596.2 volts.
Explanation:
Given that,
Capacitance
Charge
Separation of plates = 0.313 mm
We need to calculate the potential difference between the plates
Using formula of potential difference
Where, Q = charge
C = capacitance
Put the value into the formula
Hence,The potential difference between the plates is 596.2 volts.
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Answer:
D. Top is emission; bottom absorption.
Explanation:
Emission and spectrum of elements are due to the element absorbing or emitting wavelength of e-m energy. Elementary particles of elements can absorb energy from a ground state to enter an excited state, creating an absorption spectrum, or they can lose energy and fall back to a lower energy state, creating an emission spectrum. A simple rule to differentiate between an emission and an absorption spectrum is that: "all absorbed wavelength is emitted, but not all emitted wavelength is absorbed."
From the image, the lines indicates wavelengths. We can see that all of the wavelengths of the bottom absorption spectrum coincides with some of the wavelength of the upper emission wavelengths.
