00:01
In this question we have this electrical circuit and the left side is filled with distilled water.
00:09
They are both insulating containers but arrange in a like a parallel plate capacitor manner.
00:18
Okay, so the left capacity, the left thing is like a still with the water and then the right thing is still with salt water.
00:27
So there are four parts in this question.
00:30
In part a, you are given that the current is 484 milampeer, which is 0 .4 .84 ampires.
00:45
We want to find the salinity of the salt water.
00:49
So we can find the total resistance in the circuit, which is just emf divided by i10 divide by 0 .4 is 4.
01:01
This is 20 .7 oms and then the resistance of the salt water.
01:08
Okay, it's just 20 .7 minus 15.
01:12
This is 5 .66 oms actually.
01:17
And then the resistance of the salt water, you can use row l over a to represent to calculate.
01:30
And the area is pi r square okay so l here is 0 .03 meters r is 0 .05 meters okay so we can calculate and then we are given that the the resistivity of the salt water is a function of s not divided by s okay so as not is the constant salinity okay and then s is the serenity of the salt water okay in parts per 3 ,000 okay so we can calculate s and this is s not l divided by r s high r square so putting the numbers we have 6 .3 times 0 .03 divide by high by high 4 .05 square times 5 .66.
02:38
Calculate this you get 4 .25 parts per thousand.
02:48
Then in part b, we want to calculate the charge stored in the left electrical component because the whole thing acts like a capacitor.
02:59
Okay, so, and we can calculate the capacitance just like a parallel plate capacitor with a dielectric inside...