The inductor in a radio receiver carries a current of amplitude 200 mA when a voltage of amplitude 2.40 V is across it at a freq
1 answer:
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Answer:C
Explanation:
It is given that hand strap moves from the vertical in the backward direction.
The direction of strap depends upon the acceleration of bus i.e. if bus is accelerating in forward direction then strap will move in backward direction and vice-versa.
The reason for moving backwards is due to the psuedo acting on strap which bends the strap in backward direction
angle of inclination is given by
where a=acceleration of bus
=inclination of strap from vertical
so we cannot conclude anything about the direction of the velocity of the bus
Answer:
Explanation:
Solution is in the picture attached
Answer:
F= σ² L² /2ε₀
F = (L² ε₀/4π) ΔV² / r⁴
Explanation:
a) For this exercise we can use Coulomb's law
F = - k Q² / r²
where the negative sign indicates that the force is attractive and the value of the charge is equal to the two plates
Capacitance is defined by
C = Q / ΔV
Q = C ΔV
also the capacitance for a parallel plate capacitor is related to its shape
C = ε₀ A / r
we substitute
Q = ε₀ A ΔV / r
we substitute in the force equation
F = k (ε₀ A ΔV / r)² / r²
k = 1 / 4πε₀
F = ε₀ /4π L² ΔV² / r⁴4
F = L² ΔV² ε₀/ (4π r⁴)
F = (L² ε₀/4π) ΔV² / r⁴
b) Another way to solve the exercise is to use the relationship between the force and the electric field
F = q E
where we can calculate the field created by a plane using Gaussian law, where we use a cylinder with a base parallel to the plate as the Gaussian surface
Ф = ∫E .dA = / ε₀
the plate have two side
2E A = q_{int} / ε₀
E = σ / 2ε₀
σ = q_{int} / A
substituting in force
F = q σ / 2ε₀
the charge total on the other plate is
q = σ A
q = σ L²
F= σ² L² /2ε₀