If the amplitude of a wave increases by a factor of 4, how is the intensity changed?
2 answers:
The intensity increases 16 times.
The relationship between intensity and amplitude is given below:
<span> I = constant
Since, A becomes 4 times, A' = 4A</span><span>
Thus the intensity becomes, I' = constant </span>× (<span>4A)</span>²<span>
</span>⇒<span> I' = 16A</span>²
Therefore, the intensity of the wave is increased by 16 times
The relationship between intensity and amplitude is given below:
<span> I = constant
Since, A becomes 4 times, A' = 4A</span><span>
Thus the intensity becomes, I' = constant </span>× (<span>4A)</span>²<span>
</span>⇒<span> I' = 16A</span>²
Therefore, the intensity of the wave is increased by 16 times
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a) E = 0
b)
Explanation:
a)
We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:
where
E is the electric field
q is the charge contained by the Gaussian surface
is the vacuum permittivity
Here we want to find the electric field at a distance of
r = 12 cm = 0.12 m
Here we are between the inner radius and the outer radius of the shell:
However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.
This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:
q = 0
Therefore, the magnitude of the electric field is also zero:
E = 0
b)
Here we want to find the magnitude of the electric field at a distance of
r = 20 cm = 0.20 m
from the centre of the shell.
Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.
Therefore, it is given by:
where in this problem:
is the charge on the shell
is the distance from the centre of the shell
Substituting, we find: