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Define electric capacitance and write its S.I.unit.
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The ratio of charge given to a conductor and potential raised due to this charge is called electric capacitance of the conductor.
From equation :
C $$= \frac{q}{V}$$Thus , the electric capacitance of a conductor is a constant. Its value depends upon the nature and area of the conductor and on the environment around the conductor and also on the other conductors placed near the conductor.
Again from $$C$$ $$= \frac{q}{V}$$ , if $$V$$ =1 volt, then $$C = q $$
i.e. the electrical capacitance of a conductor of a conductor is equal to charge of conductor when the applied potential on a conductor take as unity.
Unit and Dimensions:
$$\because C = \frac{q}{V}$$
$$\therefore$$ Unit of capacitance $$= \frac{coulomb}{volt} =$$ CV$$^{-1} =$$ Farad (F)
$$\because$$ 1CV $$^{-1} =$$ 1F
If q = 1C, V = 1 volt, then C = 1F
i.e., 'If on giving a charge of 1 C to a conductor its potential is increased by 1 volt, then the capacitance of the conductor will be 1F'.
Farad is a big unit of capacitance. In practice other small units e.g, $$\mu$$F and $$\mu \mu F$$ or pF are used.
$$1 \mu$$F $$= 10^{-6}$$F
and $$1 \mu \mu$$F = 1 pF = $$10^{-12}$$F
Dimensional formula:
$$\because$$ $$C = \frac{q}{V} = \frac{q}{\frac{W}{q}} = \frac{q^2}{W} = \frac{i^2 t^2}{W}$$
$$\therefore$$ Dimensional formula of C $$= \frac{[A^2 T^2]}{M^1 L^2 T^{-2}} $$
$$= [M^{-1} L^{-2} T^{4} A^2]$$
Graph : If a graph is plotted between $$q$$ and $$V$$ , then the straight line is obtained.
The gradient of the graphical line $$= \tan \theta = \frac{q}{V} = C$$
i.e., the gradient of the graph provides the capacitance of the conductor.
Again from $$C$$ $$= \frac{q}{V}$$ , if $$V$$ =1 volt, then $$C = q $$
i.e. the electrical capacitance of a conductor of a conductor is equal to charge of conductor when the applied potential on a conductor take as unity.
Unit and Dimensions:
$$\because C = \frac{q}{V}$$
$$\therefore$$ Unit of capacitance $$= \frac{coulomb}{volt} =$$ CV$$^{-1} =$$ Farad (F)
$$\because$$ 1CV $$^{-1} =$$ 1F
If q = 1C, V = 1 volt, then C = 1F
i.e., 'If on giving a charge of 1 C to a conductor its potential is increased by 1 volt, then the capacitance of the conductor will be 1F'.
Farad is a big unit of capacitance. In practice other small units e.g, $$\mu$$F and $$\mu \mu F$$ or pF are used.
$$1 \mu$$F $$= 10^{-6}$$F
and $$1 \mu \mu$$F = 1 pF = $$10^{-12}$$F
Dimensional formula:
$$\because$$ $$C = \frac{q}{V} = \frac{q}{\frac{W}{q}} = \frac{q^2}{W} = \frac{i^2 t^2}{W}$$
$$\therefore$$ Dimensional formula of C $$= \frac{[A^2 T^2]}{M^1 L^2 T^{-2}} $$
$$= [M^{-1} L^{-2} T^{4} A^2]$$
Graph : If a graph is plotted between $$q$$ and $$V$$ , then the straight line is obtained.
The gradient of the graphical line $$= \tan \theta = \frac{q}{V} = C$$
i.e., the gradient of the graph provides the capacitance of the conductor.
i.e. the electrical capacitance of a conductor of a conductor is equal to charge of conductor when the applied potential on a conductor take as unity.
Unit and Dimensions:
$$\because C = \frac{q}{V}$$
$$\therefore$$ Unit of capacitance $$= \frac{coulomb}{volt} =$$ CV$$^{-1} =$$ Farad (F)
$$\because$$ 1CV $$^{-1} =$$ 1F
If q = 1C, V = 1 volt, then C = 1F
i.e., 'If on giving a charge of 1 C to a conductor its potential is increased by 1 volt, then the capacitance of the conductor will be 1F'.
Farad is a big unit of capacitance. In practice other small units e.g, $$\mu$$F and $$\mu \mu F$$ or pF are used.
$$1 \mu$$F $$= 10^{-6}$$F
and $$1 \mu \mu$$F = 1 pF = $$10^{-12}$$F
Dimensional formula:
$$\because$$ $$C = \frac{q}{V} = \frac{q}{\frac{W}{q}} = \frac{q^2}{W} = \frac{i^2 t^2}{W}$$
$$\therefore$$ Dimensional formula of C $$= \frac{[A^2 T^2]}{M^1 L^2 T^{-2}} $$
$$= [M^{-1} L^{-2} T^{4} A^2]$$
Graph : If a graph is plotted between $$q$$ and $$V$$ , then the straight line is obtained.
The gradient of the graphical line $$= \tan \theta = \frac{q}{V} = C$$
i.e., the gradient of the graph provides the capacitance of the conductor.
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