Question

Initially, the switch in FIGURE$P26.61$is in position A and capacitors $C_2$and $C_3$are uncharged. Then the switch is flipped to position $B$. Afterward, what are the charge on and the potential difference across each capacitor?

Short answer

The charge on $C_1$is $833\mu C$, and the charge on $C_2$and $C_3$is $666\mu C$. The potential difference on $C_1$is $55.5V$, the potential difference on $C_2$is $33.3V$and the potential difference on $C_3$is $22.2V$.

Step-by-step solution

Given information

We need to find the charge on and the potential difference across each capacitor.

Circuit diagram 

The equivalent circuit when the switch is in the position A is given in the figure bellow. In this case we have only one capacitor connected to the battery. Its plates are charged by where$\pm q$

$q=C_{1}V=1500\mu C$

Explanation

The piece of wire that connects with a plate of $C_1$with plate of $C_2$carries the charge $+q$.Some of this charge will be transferred to the upper plate of $C_2$and we will have

$q_1+q_2=q...\left(1\right)$

The charge of $-q_2$will be induced on the opposite plate of $C_2$. This in turn will induce $+q_2$on the neighboring plate of $C_3$(since that piece of wire is neutral in total), and the induced charge on the opposite plate will be $-q_2$. The capacitors $C_2$and $C_3$are oriented in different differently than $C_1$. Since there is no electromotive force in this circuit, the potential difference on $C_1$must be equal to the sum of potential differences on $C_2$and $C_3$i.e.

role="math" localid="1649309748138" $\frac{q_1}{C_1}=\frac{q_2}{C_2}+\frac{q_2}{C_3}...\left(2\right)$

Simplify 

Now, we have two equations for $q_1$and $q_2$. From equation $\left(1\right)$we have $q_2=q-q_1$. Substitute this into the second equation to get

$\frac{q_1}{C_1}=\frac{q-q_1}{C_2}-\frac{q-q_1}{C_3}$

This further yields

$q_1\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}=q\frac{1}{C_2}+\frac{1}{C_3}$

or

$q_1\frac{C_1{C}_2+C_2{C}_3+C_1{C}_3}{C_1{C}_2{C}_3}=q\frac{C_2+C_3}{C_2{C}_3}$

$q_1=\frac{C_1(C_2+C_3)}{C_1{C}_2+C_2{C}_3+C_1{C}_3}q=833\mu C$

and

$q_2=666\mu C$

Now, the potential differences are

$V_1=\frac{q_1}{C_1}=55.5V$.

$V_2=\frac{q_2}{C_2}=33.3V$.

$V_3=\frac{q_2}{C_3}=22.2V$.