
A $100pF$ Capacitor is charged to a potential difference of $24V$. It is connected to an uncharged capacitor of capacitance 20pF. What will be the potential difference across $100pF$ capacitor?
Answer
584.4k+ views
Hint: Capacitance is the property by virtue of which a capacitor stores charge or energy in it. In this question we need to find the total charge developed across the charged capacitor and then it is being distributed between this capacitor and uncharged capacitor according to the value of their capacitance.
The charge will be distributed until they both reach a value of common potential which is actually${{V}^{'}}$.
The distribution takes according to $Q=CV$.
Complete step by step answer:
The distribution of the total charge will take place until both capacitors reach a common potential. This takes place in accordance with conservation of charge as charge can redistribute between the charged and uncharged capacitor.
The law that relates these quantities is $Q=CV$
where$Q$ is charge $C$ is capacitance and $V$is potential difference across capacitor
Total charge on $100pF$Capacitor:
$\Rightarrow C\times V$
$\Rightarrow 100pF\times 24V$
$\Rightarrow 2400pCoulumb$
Second capacitor is uncharged so charge on it is $0$
Let the common potential across both capacitor be${{V}^{'}}$
The charge will be distributed according to the capacitance of the capacitor.
Capacitance of ${{1}^{st}}$capacitor:$100pF$
Capacitance of ${{2}^{nd}}$capacitor:$20pF$
So from conservation of charge,
Initial charge on both capacitor=final charge on both capacitor
$\Rightarrow \,\,\,2400pCoulumb=\left( 100pF+20pF \right){{V}^{'}}$
$\Rightarrow {{V}^{'}}=\dfrac{2400}{120}Volts$
$\Rightarrow {{V}^{'}}=20volts$
As it is the common potential so final voltage or potential difference across $100pF$capacitor will be $20Volts$.
Note:
The capacitance decides how much charge a capacitor will store across it.
Charge stored across capacitors is directly proportional to the capacitance of a capacitor.
And capacitance of a capacitor is inversely proportional to the potential difference applied across it.
The charge will be distributed until they both reach a value of common potential which is actually${{V}^{'}}$.
The distribution takes according to $Q=CV$.
Complete step by step answer:
The distribution of the total charge will take place until both capacitors reach a common potential. This takes place in accordance with conservation of charge as charge can redistribute between the charged and uncharged capacitor.
The law that relates these quantities is $Q=CV$
where$Q$ is charge $C$ is capacitance and $V$is potential difference across capacitor
Total charge on $100pF$Capacitor:
$\Rightarrow C\times V$
$\Rightarrow 100pF\times 24V$
$\Rightarrow 2400pCoulumb$
Second capacitor is uncharged so charge on it is $0$
Let the common potential across both capacitor be${{V}^{'}}$
The charge will be distributed according to the capacitance of the capacitor.
Capacitance of ${{1}^{st}}$capacitor:$100pF$
Capacitance of ${{2}^{nd}}$capacitor:$20pF$
So from conservation of charge,
Initial charge on both capacitor=final charge on both capacitor
$\Rightarrow \,\,\,2400pCoulumb=\left( 100pF+20pF \right){{V}^{'}}$
$\Rightarrow {{V}^{'}}=\dfrac{2400}{120}Volts$
$\Rightarrow {{V}^{'}}=20volts$
As it is the common potential so final voltage or potential difference across $100pF$capacitor will be $20Volts$.
Note:
The capacitance decides how much charge a capacitor will store across it.
Charge stored across capacitors is directly proportional to the capacitance of a capacitor.
And capacitance of a capacitor is inversely proportional to the potential difference applied across it.
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