By how much should the gap between the plates of the air capacitor be changed in the input oscillating circuit of the radio receiver in order to switch to reception of twice as long waves? The initial gap is 1 mm.
Question
By how much should the gap between the plates of the air capacitor be changed in the input oscillating circuit of the radio receiver in order to switch to reception of twice as long waves? The initial gap is 1 mm.
Solution
The frequency of a radio wave is inversely proportional to its wavelength. This means that if you want to receive a wave that is twice as long, the frequency will be half as much.
In an LC circuit, which is what the input oscillating circuit of a radio receiver essentially is, the resonant frequency is given by the formula:
f = 1 / (2π√(LC))
where L is the inductance and C is the capacitance. In this case, we are changing the capacitance by changing the gap between the plates of the air capacitor.
The capacitance of a parallel plate capacitor is given by the formula:
C = εA/d
where ε is the permittivity of the medium (air, in this case), A is the area of one of the plates, and d is the distance between the plates.
If we want to halve the frequency, we need to double the capacitance. According to the formula, we can achieve this by doubling the distance between the plates.
So, the gap between the plates should be changed from 1 mm to 2 mm.
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