I am pretty sure the answer to this is switching poles
Answer:
Electric Field = 3.369 x 10^4 N/C
Explanation:
Radius = r = (r1 + r2) / 2 = (1.6 + 3.6) /2 = 2.6 cm + 2.3 cm = 4.9 cm = 0.049 m
As we know, Electric field = E = kQ/r.r
= 8.98755 x 10^9 x 9 x 10^-9 / 0.049 x 0.049 = 33689.275 N/C
= 3.369 x 10^4 N/C
Refractive index is the ration of sin i to sin r where i is the incident angle and r is the refraction angle.
Therefore, refractive index = sin 79.5 / sin 39.6
= 1.542
The refractive index may be given by the ratio of refractive index of medium 2 to refractive index of medium 1.
Therefore, 1.542 = n/1.0003
n = 1.5425
≈ 1.54
Medium 2 is sodium chloride, refractive index of 1.54
Answer:
U/U₀ = 2
(factor of 2 i.e U = 2U₀)
Therefore, the energy stored in the capacitor is doubled when the plate separation is doubled while the capacitor has been disconnected
Explanation:
Energy stored in a capacitor can be expressed as;
U = 0.5CV^2 = Q^2/2C
And
C = ε₀ A/d
Where
C = capacitance
V = potential difference
Q = charge
A = Area of plates
d = distance between plates
So
U = Q^2/2C = dQ^2/2ε₀ A
The initial energy of the capacitor at d = d₀ is
U₀ = Q^2/2C = d₀Q^2/2ε₀ A ....1
When the plate separation is increased after the capacitor has been disconnected, the charge Q of the capacitor remain constant.
The final energy stored in the capacitor at d = 2d₀ is
U = 2d₀Q^2/2ε₀ A ...2
The factor U/U₀ can be derived by substituting equation 1 and 2
U/U₀ = (2d₀Q^2/2ε₀ A)/( d₀Q^2/2ε₀ A )
Simplifying we have;
U/U₀ = 2
U = 2U₀
Therefore, the energy stored in the capacitor is doubled when the plate separation is doubled while the capacitor has been disconnected.
Time in the universe vs the population