00:01
In this particular question we have given a parallel plate capacitors with its capacitance, its area, the voltage which is applied on it, and we have to calculate the electric field, the free charge and bound charge for this particular parallel plate capacitor.
00:15
So first let me just write the values of the quantities which we have given in this particular question.
00:21
The capacitance is given as 100 picof a ferr.
00:26
The area of the plates is given as 100 cm squares.
00:30
The k value for the dielectric in between the plates of the parallel plate capacitor is given as 5 and the voltage which is being applied on the plates is given as 25 volts.
00:47
So first let's just convert this quantity, the value of these quantities into si units.
00:53
So for capacitance we can write it as 100 multiplied by 10 to power minus 12 ferrar.
01:01
So what i have done is just i have converted it into si units for area i can write it is hundred multiplied by tenor's to power minus four meter square so i have converted the centimeter squares into meter squares and now the k value and the voltage value are already given in the si units so we don't have to do anything with them so let's just move to the a part of this question in the a part of this question we have to calculate the electric field in between the plates and another thing which is missing here is the distance of separation of the plates so i can calculate the distance of separation of the plates using the formula for the capacitance so i can write the capacitance is area time epsilon 0 times the k divided by distance so from here i can make the d as a subject and i can write it as a epsilon 0 times k divided by now let's just look at the a part of this question in the a part of this question we have to calculate the electric field the electric field between the plates so let's just write the electric field sv over d the voltage divided by the distance of separation and i can replace the d with c divided by a absalom or k so i just replace the d using this formula this above formula now i can put all the values the value of the v is given already to us which is 25 times the value of the capacitance which is given as 100 times 10 to power minus 12 in si units divided by the area which is given as 100 cm square so i can write that in si units as this the epsilon 0 is a constant value which is 8 .85 it's a constant value it's a constant value for free space which is 8 .85 times 10x to power minus 12 and the value is again in assign it so we don't have to worry about the units because whatever answer we will get that will be newton per column and if you do all this calculation you will get for the electric field the value will be 5649 .7 newton per column this will be the electric field between the plates of the capacitor now let's just move to the b part of this question where we have to calculate the free charge on the plates of the capacitor so we know that the free charge is equals to c times v c is the capacitance of the capacitor and v is the voltage which is being applied on it so the values are already given to us the value of the c is 100 times tender to power minus 12 and the value of the v is quantified so if you do this simple calculation you will get 2 .5 nanoculum.
04:14
Now let's just move to the third part of this question...