You want the current amplitude through a 0.450-mH inductor (part of the circuitry for a radio receiver) to be 1.90 mA when a sin
1 answer:
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Explanation:
Formula for angle subtended at the center of the circular arc is as follows.
where, S = length of the rod
r = radius
Putting the given values into the above formula as follows.
=
=
=
Now, we will calculate the charge density as follows.
=
=
Now, at the center of arc we will calculate the electric field as follows.
E =
=
= 34.08 N/C
Thus, we can conclude that the magnitude of the electric eld at the center of curvature of the arc is 34.08 N/C.
Answer:
Will be doubled.
Explanation:
For a capacitor of parallel plates of area A, separated by a distance d, such that the charges in the plates are Q and -Q, the capacitance is written as:
where e₀ is a constant, the electric permittivity.
Now we can isolate V, the potential difference between the plates as:
Now, notice that the separation between the plates is in the numerator.
Thus, if we double the distance we will get a new potential difference V', such that:
So, if we double the distance between the plates, the potential difference will also be doubled.
Answer:
Explanation:
Attach is the solution
