Question

find the charge on each capacitor ​

Answer 1

Answer:

6μF capacitor = 100μC | 2μF capacitor = 100μC | 1μF capacitor = $\frac{200}{3}$μC

Explanation:

When doing this sum first you must know some things about how capacitor connections effect the amount of charges stored in a capacitor,

The capacitors which are connected in series function kind of differently to the capacitors which are connected in parallel. In this problem as you can see the 2μF and 6μF capacitors are connected in series to each other while the 1μF capacitor is connected parallel to other two capacitors so you have to deal with them separately.  

When using capacitance equation,

For the capacitors connected in parallel you can directly use the equation since they are directly connected to the battery.

But for the capacitors which are connected in series you have to find the equivalent capacitance before using the capacitance equation to find the charge stored in it.

That is the basic idea I will explain further as we progress through the sum.

So lets get into the circuit,

First you need to find the potential at the A (Simply the potential drop across the 20Ω resistor)

[Hope you know the reason for finding the potential at A,in case if you don't please make sure to ask.]

To find that you can simply use the voltage divider method since the two resistors are connected in series.

 So the potential at A will be  

  (20Ω/20Ω+10Ω) × 100V = 200/3V

Now we are really going to start calculating the charge in each capacitor.

I think you already know the equation to find the capacitance of a capacitor which is $C = \frac{Q}{V}$

Consider the 1μF capacitor which is connected parallel (I mean parallel to the other two capacitors, which means it is directly connected to the 200/3V potential)

Here just substitute the values to Q=CV as I said above

So the charge in the capacitor 1μF will be,

Q = CV (from the equation C=Q/V)

Q = 1μF * 200/3V

therefore, Q = 200/3μC

Now lets get to the capacitors in series,

Here first you need to find the equivalent capacitance which can be taken as,

$\frac{1}{C_e}= \frac{1}{C_{1} }+\frac{1}{C_{2} }...$

Equivalent capacitance in this case can be taken as  

1/Ce = (1/2 + 1/6)

1/Ce = 3+1/6

therefore, Ce = 6/4μF

Now substitute the obtained equivalent resistance in the equation Q=CV to find the charge in the capacitor,

Q = CV

Q = 6/4μF × 200/3V

Q = 100μC

The amount of charges we found above is amount of charges stored in each of the two capacitors as they are connected in series with each other(2μF and 6μF capacitors, and yes the amount of charges stored in each capacitor is the same).  

So lastly I would like to teach you something about the capacitors in series, the amount of charges stored in each capacitor in a network or a circuit where capacitors are connected in series to each other is always the same in all capacitors irrespective of their capacitance.

Please make sure to ask if you have any questions. Hope I made no mistakes ;)