
How big is a 1-farad capacitor?
Answer
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Hint: A capacitor is a device that stores the energy When an insulated conductor is given a charge it develops a certain voltage. As we increase the charge the potential developed also increases. The ratio of charge and voltage is called capacitance. Capacitance is the ability of the capacitor to hold the charge.
Complete step by step answer:
As we know the capacitance is the ratio of charge and voltage. We can write it mathematically, as below.
\[ \Rightarrow C = \dfrac{Q}{V}\]
Here, $C$ is the capacitance, $Q$ is the charge and $V$ is the potential.
The SI unit of capacitance is Farad (F). A conductor has one-farad capacitance if adding one coulomb of charge increases its voltage by one volt.
\[ \Rightarrow C = \dfrac{{1Coulomb}}{{1volt}}\]
Subsitute the value of one coulomb.
\[ \Rightarrow C = \dfrac{{6.25 \times {{10}^{18}}electrons}}{{1volt}}\]
Hence, One-farad capacitance is obtained if $6.25 \times {10^{18}}$ electrons are transferred from one plate to another due to this one-volt voltage being increased.
Note:
One farad-capacitance is a large value, practically it's not possible.
A positively charged plate is placed with an uncharged plate. Due to induction, the negative charge develops on the face closer to the first plate and the positive charge develops on the face farther from the first plate.
The negative charge developed tries to decreases the potential of the first plate and the positive charge on the second plate tries to increase the potential on the first plate. But because the negative is closer to the first plate hence, the potential of the first plate decreases by a small amount.
We know voltage is inversely proportional to the capacitance, hence as voltage decreases the capacitance of the capacitor increase.
Complete step by step answer:
As we know the capacitance is the ratio of charge and voltage. We can write it mathematically, as below.
\[ \Rightarrow C = \dfrac{Q}{V}\]
Here, $C$ is the capacitance, $Q$ is the charge and $V$ is the potential.
The SI unit of capacitance is Farad (F). A conductor has one-farad capacitance if adding one coulomb of charge increases its voltage by one volt.
\[ \Rightarrow C = \dfrac{{1Coulomb}}{{1volt}}\]
Subsitute the value of one coulomb.
\[ \Rightarrow C = \dfrac{{6.25 \times {{10}^{18}}electrons}}{{1volt}}\]
Hence, One-farad capacitance is obtained if $6.25 \times {10^{18}}$ electrons are transferred from one plate to another due to this one-volt voltage being increased.
Note:
One farad-capacitance is a large value, practically it's not possible.
A positively charged plate is placed with an uncharged plate. Due to induction, the negative charge develops on the face closer to the first plate and the positive charge develops on the face farther from the first plate.
The negative charge developed tries to decreases the potential of the first plate and the positive charge on the second plate tries to increase the potential on the first plate. But because the negative is closer to the first plate hence, the potential of the first plate decreases by a small amount.
We know voltage is inversely proportional to the capacitance, hence as voltage decreases the capacitance of the capacitor increase.
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