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Consider the following problem:

There are two spherical conductors $A, B$, with capacitances $C_A, C_B$ resp. Conductor $A$ is supplied some charge and is found to have a potential of $160 \space V$. Conductor $B$ is also supplied some charge and is found to have a potential of $50 \space V$. They are then connected so that their charges equalize. Now, their common voltage is $140 \space V$. Find $C_A: C_B$.

I started off with the basic relation, $Q_A = C_AV_A = 160\space C_A$ similarly, $Q_B = 50C_B$. Now, when they are connected the total charge will equal $Q_T = Q_A + Q_B = 160C_A + 50C_B$. Now, the total capacitance of the combination of these spheres will be $C_T$.

I can't seem to decide if these spheres should be considered to be connected in parallel or in series.

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1 Answer 1

You don't need to determine whether they are in series or parallel to solve this problem(They might not be in either).

The final voltage of both spheres is given to you, so you can find the charges which will finally appear on both the spheres, given by

$$Q_f=140C_A+140C_B$$

This must be enough for you to solve further. If not then hover over the boxes below to see further solution

By conservation of charge we have

$$Q_f=Q_{A,i}+Q_{B,i}$$
$$140C_A+140C_B= 160C_A + 50C_B$$

Thus

$$90C_B=20C_A$$
$$\therefore \frac{C_A}{C_B}=\frac 92$$

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