11. Two radio antennas are 120 m apart on a north-south line, and they radiate in phase at a frequency of 3.4 MHz. All radio mea
1 answer:
Answer:
47°
Explanation:
between the two antennas for an observer difference in length in quadrant I is
ΔL = dCos(θ)
setting θ=0 i.e the observer is north of the station to get
ΔL= d= 120 m.
The wavelength formula is:
λ = c / f = 3x / 3.4 x
= 88.2m
Constructive interference occurs when the Length difference is equal to an integer multiple of a wavelength.
At the given spacing, and from an observation point north of the stations, they are more than a wavelength apart, so they are not in constructive interference. Looking for the first peak:
dCos(θ) = λ
Cos(θ) = λ/d
θ = arcCos(λ/d) = arcCos( 88.2 / 120 ) = 42.7°= 43°
For destructive interference:
90° - 43°= 47°
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Answer:
The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)
Explanation:
The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other. , the amount of charge stored in this capacitor, will stay the same.
The formula relates the electric potential across a capacitor to:
, the charge stored in the capacitor, and
, the capacitance of this capacitor.
While stays the same, moving the two plates apart could affect the potential
by changing the capacitance
of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:
,
where
is the permittivity of the material between the two plates.
is the area of each of the two plates.
is the distance between the two plates.
Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of . Neither will that change the area of the two plates.
However, as (the distance between the two plates) increases, the value of
will become smaller. In other words, moving the two plates of a parallel-plate capacitor apart would reduce its capacitance.
On the other hand, the formula can be rewritten as:
.
The value of (charge stored in this capacitor) stays the same. As the value of
becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.