91Ó°ÊÓ

The capacitor which is connected parallel with C when C is fully charged.

$$\begin{gathered} a)C\text{'}=60×{10}^{-6}F \\ b)V=100V \end{gathered}$$

The law of conservation of the charge states that the total charges are conserved, we cannot create or destroy the charges. The charge on the capacitor is equal to the product of capacitance and voltage of the capacitor.

We will use conservation of charge, in this case, to find the capacitance.

Formulae:

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

According to the conservation of charge,

$Q=q_1+q_2$ (2)

Let Q be the charge on the capacitor C when it is fully charged to potential V = 100 V

This equation implies the following condition derived from the formula of equations (i) and (ii) as:

$$\begin{gathered} \mathrm{C}×100\mathrm{V}=\mathrm{C}×40\mathrm{V}+\mathrm{C}\text{'}×40\mathrm{V} \\ \mathrm{C}×(100-40)=\mathrm{C}\text{'}×40 \\ \mathrm{C}×60=×60×{10}^{-6}\mathrm{F}×40 \\ \mathrm{C}=\frac{60×{10}^{-6}\mathrm{F}×40}{60} \\ =40×{10}^{-6}\mathrm{F}\mathrm{or}40\mathrm{μ¹ó} \end{gathered}$$

Hence, the value of the capacitance is $40\mathrm{μ¹ó}$.