Question

How to solve? Help please!

Answer 1

`"See the description section for details."`

Explanation:

• First we must calculate the system's capacity.
• The capacitors are connected in series.
• We can calculate the equivalent capacitance of series-connected capacitors using the formula given below.

`C=(C_1*C_2)/(C_1+C_2)`

`C=(6*3)/(6+3)=18/9=2 mu F`

`2 mu F=2*10^-6 F`

• Let's calculate the electrical charge stored in the system.

`Q=C*V`

`"where;"`

`Q:"charge stored in system"`

`V:10V`

`C=2*10^-6 F`

`Q=2*10^-6*10=2*10^-5 C`

• The amount of charge stored by the series connected capacitors is equal.
• The electric charge of each of the series-connected capacitors is equal to the electric charge of the system.

`"charge stored on "C_1:2*10^-5 C`

`"charge stored on "C_2:2*10^-5 C`

`"charge stored on system "C:2*10^-5 C`

• In the second circuit we see that the poles have changed. The capacitors are connected in parallel.

• Since the voltages of parallel-connected capacitors are equal, it is necessary to make a charge transfer between the capacitors.

-The new charge distribution is as follows.

`"if charge of "C_2=x`

`"then charge of "C_1=2x`

`x+2x="system's charge"`

`3x=2*10^-5`

`x=2/3*10^-5C" charge of "C_2`

`2x=2*2/3*10^-5=4/3 *10^-5C" charge of C_1"`