In a RC circuit, when a capacitor is discharging, the charges, current and voltage across the capacitor are decreasing exponentially. It's a continuous process. The bigger the R, the slower the discharing process.
Finally I ended up with this equation:
\kappa=\frac{Q}{Q-\epsilon_{0}AE_D}
where E_D denotes the electric field within the diaeletric. Am I right? Btw, what's the answer?
As refraction index n=\frac{c}{v} suggests that all mediums other than vacuum has n>1. I would like to know that is there any material that allows light to gain more speed rather than losing speed? If no, can it be proved? Why you are so sure of your answer?
I just had my physics test today. It's a nightmare. I didn't really have time to finish them all. Those questions are not very difficult except for some but the time is very limited. And my style for doing questions is not to recall formulas Rather, I would derive them. Test should be testing us...
I think the question is like this.
First the Q is transferred from the 394K reservoir to the reservoir with temperature of T. This Q will then become the 100% of the input energy that works between the T and the 248K reservoir. When we want to find T, there are 2 possibility for T, either it is...
Here is my working,
Efficiency = \frac{output work}{input heat}
thus Efficiency = 0.7
For Carnot engine,
Efficiency = 1 - \frac{T_c}{T_h}
The T_c is the T that we want to find. Why we can't put it as a denominator? Well, the first info tells us that it's not possible.
As for your...
Vincent, I think I know my problem. The circuit is considered as if C_1 is connected parallelly to C_2 and C_3 which are in series. Therefore, it is reasonable for the 6\mu F capacitor to have smaller charge compare to the 4\mu F capacitor because it is in a series which is parallel to the 4\mu...
R_1 = 0.2m, R_2 = 0.3m, R_3 = 0.4m
For r < R_1,
Since the solid sphere is a conductor, the E field will be 0.
For R_1 < r < R_2,
\phi = \frac{q_{in}}{\varepsilon}
\phi = E\times 4\Pi r^2
So E = ...
For R_2 < r < R_3,
It's same with first one.
For r > R_3,
Use the method as for R_1 < r...
Ok, now I work out what I understood from your solution,
C_1 = 4\mu F, C_2 = 3\mu F, C_3 = 6\mu F
So, C_4 = (C_{2}^{-1} + C_{3}^{-1})^{-1} = 2\mu F
\frac{Q_1}{C_1} = \frac{Q_4}{C_4}
\Rightarrow Q_1 = 2Q_4 ---- 1
Q_1 + Q_4 = 48\mu C ---- 2
2Q_4 + Q_4 = 48\mu C
Q_4 = 16\mu C
Q_1 = 48 - 16...
I think eventually they will all have the same amount of charges because they are connected in series. What will actually change is the final voltage.
My approach is as follow:
Q = C_{1}V_1 = 48\mu C
C_4 = (C_{1}^{-1} + C_{2}^{-1} +C_{3}^{-1})^{-1}
C_4 = \frac{4}{3} \mu F
C_1 V_1 = C_4 V_4...