Answer: 0.93 mA
Explanation:
In order to calculate the current passing through the water layer, as we have the potential difference between the ends of the string as a given, assuming that we can apply Ohm’s law, we need to calculate the resistance of the water layer.
We can express the resistance as follows:
R = ρ.L/A
In order to calculate the area A, we can assume that the string is a cylinder with a circular cross-section, so the Area of the water layer can be written as follows:
A= π(r22 – r12) = π( (0.0025)2-(0.002)2 ) m2 = 7.07 . 10-6 m2
Replacing by the values, we get R as follows:
R = 1.4 1010 Ω
Applying Ohm’s Law, and solving for the current I:
I = V/R = 130 106 V / 1.4 1010 Ω = 0.93 mA
Answer: The electric field decreases because of the insertion of the Teflon.
Explanation:
If the charge on the capacitor is held fixed, the electric field as a consequence of this charge distribution (directed from the positive charged plate to the negative charged one remains unchanged.
However, as the Teflon is a dielectric material, even though doesn't allow the free movement of the electrons as an answer to an applied electric field, it allows that the electrons be displaced from the equilibrium position, leaving a local negative-charged zone close to the posiitive plate of the capacitor, and an equal but opposite charged layer close to the negative plate.
In this way, a internal electric field is created, that opposes to the external one due to the capacitor, which overall effect is diminishing the total electric field, reducing the voltage between the plates, and increasing the capacitance proportionally to the dielectric constant of the Teflon.
Answer:
The solution for the given problem is done below.
Explanation:
M1 = 2.0
= 0.3636
= 0.5289
= 0.7934
Isentropic Flow Chart: M1 = 2.0 , 
= 1.8
T1 = 
(1.8)(288K) = 653.4 K.
In order to choke the flow at the exit (M2=1), the above T0* must be stagnation temperature at the exit.
At the inlet,
T02= 
= (1.8)(288K) = 518.4 K.
Q= Cp(T02-T01) = 
= 135.7*
J/Kg.
