91Ó°ÊÓ

1. Capacitance of the capacitor 1,$C_1=6.00\mathrm{μ}F$
2. Capacitance of capacitor 2,$C_2=4.00\mathrm{μ}F$

c. Potential difference across the pair, V = 200 V

We first find the equivalent capacitance of the series combination and then using the relation of charge and capacitance we find the charge on each capacitor and the potential difference across it.

Formulae:

The equivalent capacitance of a series connection of capacitors,

$\frac{1}{C_{equivalent}}=\underset{}{∑}\frac{1}{C_i}$ …(¾±).

The charge stored between the plates of the capacitor, q=CV …(ii)

Since two capacitors are in series their equivalent capacitance is given using equation (i) by,

$$\begin{gathered} \frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2} \\ {C}_{eq}=\frac{C_1{C}_2}{C_1+C_2} \\ =\frac{6\mathrm{μ}F×4\mathrm{μ}F}{6\mathrm{μ}F+4\mathrm{μ}F} \\ =2.40\mathrm{μ}F \end{gathered}$$

Hence, the value of the capacitance is $2.40\mathrm{μ}F$.

The value of the charge using the potential difference across the battery can be given using equation (ii) as:

$\begin{array}{ll}{q}_1=2.40\mathrm{μ}F×200& V\\ =480\mathrm{μ}C& \\ =4.80×{10}^{-4}C& \end{array}$

Hence, the value of the charge is $4.80×{10}^{-4}C$.

The value of the potential acrossis given using the given data in equation (ii) as follows:

$\begin{array}{l}{V}_1=\frac{4.80×{10}^{-4}C}{6.00\mathrm{μ}{F}^F}\\ =80.0V\end{array}$

Hence, the value of the potential is 80.0 V.

Since the two capacitors are in series the charge across each capacitor is same. Hence the charge on $C_1$and $C_2$is same, which is $4.80×{10}^{-4}C$.

Total potential difference V is the sum of the potential difference across $C_1$and$C_2$.

Thus, the value of the potential difference$V_2$ is given using the data in equation (ii) as follows:

$\begin{array}{l}{V}_2=V-V_1\\ =200V-80.0V\\ =120V\end{array}$

Hence, the value of the potential difference is 120 V.