Question

Capacitors $C_1=10\mu \mathrm{F}$and $C_2=20\mu \mathrm{F}$are each charged to $10\mathrm{V}$, then disconnected from the battery without changing the charge on the capacitor plates. The two capacitors are then connected in parallel, with the positive plate of $C_1$connected to the negative plate of $C_2$and vice versa. Afterward, what are the charge on and the potential difference across each capacitor?

Short answer

The charge on $C_1$ is$33\mu \mathrm{C}$ and the charge on $C_2$ is $66\mu \mathrm{C}$. The potential difference is $3.33V$ on both of them.

Step-by-step solution

Given Information

Capacitor$C_1=10\mu \mathrm{F}$

Capacitor$C_2=20\mu F$

Potential energylocalid="1648547398254" $=10V$

Explanation

Capacitors are connected to the battery,

$Q_1=C_{1}V=100\mu \mathrm{C}$

and

$Q_2=C_{2}V=200\mu \mathrm{C}$

During the connecting process, since you are connecting plates with opposite signs of charge, the wire connecting them will carry the charge of $Q_2-Q_1=100\mu \mathrm{a}$and $Q_1-Q_2=-100\mu \mathrm{C}$.

Since the total charge of the wire gets redistributed, the plates of the two capacitors now carry some new charges from $q_1$and $q_2$.

The previously positive plate of$C_2$stays positive and that previously positive plate of $C_1$now becomes negative.

This is because, after redistribution, neighboring plates have the same sign of charge.

The charge conservation implies,

$q_1+q_2=Q_2-Q_1=100\mu \mathrm{C}$

Another equation for $q_1$and $q_2$is found from the fact that, since there is no emf in this circuit, the potential difference must be the same on both capacitors:

$\frac{q_1}{C_1}=\frac{q_2}{C_2}$

This seconds equation yields

$q_2=\frac{C_2}{C_1}{q}_1=2q_1$

So, the expression will be

$3q_1=100\mu \mathrm{C}$

We will get,

$q_1=33\mu \mathrm{C}$

And thus

$q_2=67\mu \mathrm{C}$

The potential difference is the same and equal to

localid="1648621897464" $V=\frac{q_1}{C_1}$

Substitute the values,

localid="1648824204412" $V=\left(\frac{33J}{10\mu F}\right)=3.3V$

Hence, The charge on localid="1648824211799" $33\mu C$and the charge on localid="1648824219418" $C_2$is localid="1648824225536" $66\mu C.$The potential difference is$3.33V$on both of them.