Answer:
1 × 10⁶ N/C
Explanation:
The magnitude of the electric field between the membrane = surface density / permittivity of free space = 10 ⁻⁵C/ m² / (8.85 × 10⁻¹²N⁻¹m⁻²C²) = 1.13 × 10⁶ N/C approx 1 × 10⁶ N/C
D. frequency of the corresponding light wave
The amplitude of a wave can be obtained by measuring the distance from the resting position of the wave to its crest. The resting position is half of the distance from the crest to the trough. Given that the distance between the crest and the trough is 3 meters, the amplitude should be half of that, which is 1.5 meters.
Answer:
ε₂ =2.63 V
Explanation:
given,
M = 0.0034 H
I (t) = I₀ sin (ωt)
I (t) = 5.4 sin (143 t)
magnitude of the induced emf in the second coil
ε₂ =
ε₂ =
for maximum emf
cos (143 t) = 1
ε₂ =
ε₂ =2.63 V
Answer:
i E=V/d=50/2*10^-3=25*10^3 N/C
ii It is a (+) and (-)
iii C=εA/d
C=12.56*10^-8 * 0.1/2*10^-4
C=62.83 μF
Q=CV=50*6.283*10^-6
Q=314 μC
iv E=0.5 QV
=0.5(50*314*10^-6)
=7850 μJoule
Explanation:
