Question 78: Electrostatics & Capacitance

Question

A capacitor $P$ with capacitance $10 	imes 10^{-6} \mathrm{~F}$ is fully charged with a potential difference of 6.0 V and disconnected from the battery. The charged capacitor $P$ is connected across another capacitor $Q$ with capacitance $20 	imes 10^{-6} \mathrm{~F}$. The charge on capacitor $Q$ when equilibrium is established will be $\alpha 	imes 10^{-5} C$ (assume capacitor $Q$ does not have any charge initially), the value of $\alpha$ is $\_\_\_\_$ .

Accepted Answer

When capacitor P is fully charged by the 6.0 V battery, it stores an initial amount of charge ( $\mathrm{Q}_{	ext {initial }}$ ).
$$ Q=C \cdot V $$
Given the values for capacitor $P$ the capacitance is $C_P=10 	imes 10^{-6} \mathrm{~F}$
$$ Q_{	ext {initial }}=\left(10 	imes 10^{-6}ight) 	imes 6.0 \mathrm{C} $$
$$\Rightarrow $$ $$ Q_{	ext {initial }}=60 	imes 10^{-6} \mathrm{C} $$
After being disconnected from the battery, capacitor $P$ is connected across an initially uncharged capacitor $Q$.
The charges flow from P to Q until both capacitors reach the same electrical potential.
According to the law of conservation of charge, the total charge in the system remains constant :
$$ Q_{	ext {total }}=Q_{	ext {initial }}=60 	imes 10^{-6} \mathrm{C} $$
In a parallel connection, the equivalent capacitance ( $\mathrm{C}_{\mathrm{eq}}$ ) is the sum of the individual capacitances :
$$ C_{\mathrm{eq}}=C_{\mathrm{P}}+C_{\mathrm{Q}} $$
$$\Rightarrow $$ $C_{\mathrm{eq}}=\left(10 	imes 10^{-6}ight)+\left(20 	imes 10^{-6}ight)$
$$\Rightarrow $$ $C_{\mathrm{eq}}=30 	imes 10^{-6} \mathrm{~F}$
$$\Rightarrow $$ $\mathrm{V}_{	ext {common }}=\frac{\mathrm{Q}_{	ext {total }}}{\mathrm{C}_{\mathrm{eq}}}$
$$\Rightarrow $$ $$ \mathrm{V}_{	ext {common }}=\frac{60 	imes 10^{-6}}{30 	imes 10^{-6}} \mathrm{~V}=2.0 \mathrm{~V} $$
The final charge $\left(\mathrm{Q}_{\mathrm{Q}}ight)$ using the standard capacitor formula :
$$ \mathrm{Q}_{\mathrm{Q}}=\mathrm{C}_{\mathrm{Q}} \cdot \mathrm{~V}_{	ext {common }} $$
$$\Rightarrow $$ $\mathrm{Q}_{\mathrm{Q}}=\left(20 	imes 10^{-6}ight) 	imes 2.0 \mathrm{C}$
$$\Rightarrow $$ $$ \mathrm{Q}_{\mathrm{Q}}=40 	imes 10^{-6} \mathrm{C} $$
So, the charge on capacitor Q will be $4 	imes 10^{-5} \mathrm{C}$.
$$ \alpha 	imes 10^{-5} \mathrm{C}=4.0 	imes 10^{-5} \mathrm{C} \Rightarrow \alpha=4 $$

Electrostatics & Capacitance

Integer / Numeric Answer Type