When the top right-hand capacitor is filled with a material of dielectric constant κ, the charge on this capacitor is increases
by a factor of 1.35. Find the dielectric constant κ of the material inserted into the top right-hand capacitor.
2 answers:
Answer:
Explanation:
Given:
Capacitance, c1 = c2 = c3 = c4
= 80 micro farad
Voltage, v = Eb
= 4 V
We are not concerned with c3 and c4 just c1 and c2,
Since c1 and c2 are in series,
1/ct = 1/c1 + 1/c2
vab = v1 + v2
= q'/c1 + q'/c2
But c2 = κ × c,
= q'/c + q'/κc .....1
Let c1 = c2 = c,
Vab = q/c + q/c
= 2 × q/c. ........2
Equating equation 1 and 2,
q'/c + q'/κc = 2q/c
q'/c × (1 + 1/κ) = 2q/c
q'/q = 2κ/(1 + κ)
But q'/q = 1.35
2κ/(1 + κ) = 1.35
0.65κ = 1.35
κ = 1.35/0.65
= 2.077
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Answer:
a)
b)
Explanation:
a) Let's use the constant velocity equation:
- v is the speed of the muon. 0.9*c
- c is the speed of light 3*10⁸ m/s
b) Here we need to use Lorentz factor because the speed of the muon is relativistic. Hence the time in the rest frame is the product of the Lorentz factor times the time in the inertial frame.
v is the speed of muon (0.9c)
Therefore the time in the rest frame will be:
No we use the value of Δt calculated in a)
I hope it helps you!