91Ó°ÊÓ

a) Capacitance,$C_1=6.00\mathrm{μ¹ó}$

b) Capacitance,$C_2=4.00\mathrm{μ¹ó}$

c) Potential of battery,$V=200V$

d) Here given capacitors in parallel.

We first find the equivalent capacitance of the parallel 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 parallel connection of capacitors,

$C_{equivalent}=∑C_i$ …(¾±)

The charge stored between the plates of the capacitor,$q=CV$ …(¾±¾±)

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

$$\begin{gathered} C_{eq}=6.00\mathrm{μ}F+4.00\mathrm{μ¹ó} \\ =10.0\mathrm{μ¹ó} \end{gathered}$$

Hence, the value of the capacitance is $10.0\mathrm{μ¹ó}$.

Since the two capacitors are now parallel, the potential difference across both the capacitors is same but charge is different.

Thus, the value of the charge of capacitor 1 is given using equation (ii) as:

$$\begin{gathered} q_1=6.00\mathrm{μ}F×200V \\ =1200\mathrm{μ}C \\ =1.2×{10}^{-3}C \end{gathered}$$

Hence, the value of the charge is $1.20×{10}^{-3}C$.

Since the two capacitors are now parallel, the potential difference across both the capacitors is same. That is given by,

$$\begin{gathered} V_1=V \\ =200V \end{gathered}$$

Hence, the value of the potential difference is 200V.

The charge on$C_2$ is given using the data in equation (ii) as follows:

$$\begin{gathered} q_2=4.00\mathrm{μ}F×200V \\ =8×{10}^{-4}C \end{gathered}$$

Hence, the value of the charge is $8×{10}^{-4}C$.

Since the two capacitors are now parallel the potential difference across both the capacitors is same.

Hence, the value of the potential difference across capacitor 2 is 200V.