How is the electrostatic force affected when the magnitude of a charge is doubled?
1 answer:
The magnitude of the electrostatic force between two charges is given by:
where
ke is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges
We can see that the magnitude of the force is directly proportional to the charges. This means that when one of the charges is doubled, the magnitude of the electrostatic force will double as well, so the correct answer is
A) <span>The magnitude of the electrostatic force doubles</span>
where
ke is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges
We can see that the magnitude of the force is directly proportional to the charges. This means that when one of the charges is doubled, the magnitude of the electrostatic force will double as well, so the correct answer is
A) <span>The magnitude of the electrostatic force doubles</span>
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Answer:
z = 0.8 (approx)
Explanation:
given,
Amplitude of 1 GHz incident wave in air = 20 V/m
Water has,
μr = 1
at 1 GHz, r = 80 and σ = 1 S/m.
depth of water when amplitude is down to 1 μV/m
Intrinsic impedance of air = 120 π Ω
Intrinsic impedance of water =
Using equation to solve the problem
E(z) is the amplitude under water at z depth
E_o is the amplitude of wave on the surface of water
z is the depth under water
now ,
taking ln both side
21.07 x z = 16.81
z = 0.797
z = 0.8 (approx)
The frequency of the wave is
Explanation:
The frequency, the wavelength and the speed of a wave are related by the following equation:
where
c is the speed of the wave
f is the frequency
is the wavelength
For the radio wave in this problem,
Therefore, the frequency is:
Learn more about waves here:
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