Question
Write the definition of electric capacitance. A parallel plate capacitor is filled with a medium of dielectric constant K between the plates. Find the expression of its capacitance. Draw the necessary diagram.
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Answer
When an isolated conductor is charged, its potential increases. The increase in potential is proportional to the charge supplied to the conductor.
$Q \propto V$ or $Q=C V$
Here the proportionality constant C is called the capacitance of the conductor. The value of capacitance C of a conductor depends on its size, shape and the medium used in its construction. Keep in mind that the value of capacitance of a conductor does not depend on the charge on the conductor and the increase in potential across it. If V $=1$ volt in equation (1), then $Q = C$ i.e., the capacitance of the conductor is equivalent to the amount of charge which increases the potential of the conductor by unit amount.
The S.I. unit of capacitance is coulomb/volt, which is called Farad. If in equation (1) $Q =1$ coulomb, $V =1$ volt then $C =1$ Farad.i.e., if on giving 1 coulomb of charge to a conductor, its potential increases by 1 volt, then the capacitance of that conductor is called 1 Farad.
Parallel Plate Capacitor Completely Filled with a Dielectric Here the picture shows a parallel plate capacitor completely filled with dielectric material. The dielectric constant of the material is K and the area of the plates of the capacitor is A and the distance between them is $d$. If in the absence of dielectric material, the charge on the capacitor is $Q _0$ and the potential difference between the plates is $V _0$, then the capacitance of the capacitor in air or vacuum will be $ C_0=\frac{Q_0}{V_0} $
When a material of relative dielectric constant K is filled between the plates of a capacitor, the value of charge on the capacitor does not change due to the presence of the dielectric material filled between the plates. Rather, the value of the potential difference between the plates decrease by $\left(\frac{1}{K}\right)$ times. Therefore potential difference between the plates
$ V=\frac{V_0}{K} $
If the value of capacitance of a capacitor filled with dielectric constant is C then
$C=\frac{Q_0}{V}=\frac{Q_0}{V_0 / K}$
$C =\frac{ Q _0 K}{ V _0}$
$C = C _0 K$
or $C = K \left(\frac{\epsilon_0 A}{d}\right)$
Therefore, the value of capacitance C of a parallel plate capacitor completely filled with dielectric material is K times the value of air or vacuum capacitance $C_0$.
$Q \propto V$ or $Q=C V$
Here the proportionality constant C is called the capacitance of the conductor. The value of capacitance C of a conductor depends on its size, shape and the medium used in its construction. Keep in mind that the value of capacitance of a conductor does not depend on the charge on the conductor and the increase in potential across it. If V $=1$ volt in equation (1), then $Q = C$ i.e., the capacitance of the conductor is equivalent to the amount of charge which increases the potential of the conductor by unit amount.
The S.I. unit of capacitance is coulomb/volt, which is called Farad. If in equation (1) $Q =1$ coulomb, $V =1$ volt then $C =1$ Farad.i.e., if on giving 1 coulomb of charge to a conductor, its potential increases by 1 volt, then the capacitance of that conductor is called 1 Farad.
Parallel Plate Capacitor Completely Filled with a Dielectric Here the picture shows a parallel plate capacitor completely filled with dielectric material. The dielectric constant of the material is K and the area of the plates of the capacitor is A and the distance between them is $d$. If in the absence of dielectric material, the charge on the capacitor is $Q _0$ and the potential difference between the plates is $V _0$, then the capacitance of the capacitor in air or vacuum will be $ C_0=\frac{Q_0}{V_0} $
When a material of relative dielectric constant K is filled between the plates of a capacitor, the value of charge on the capacitor does not change due to the presence of the dielectric material filled between the plates. Rather, the value of the potential difference between the plates decrease by $\left(\frac{1}{K}\right)$ times. Therefore potential difference between the plates
$ V=\frac{V_0}{K} $
If the value of capacitance of a capacitor filled with dielectric constant is C then
$C=\frac{Q_0}{V}=\frac{Q_0}{V_0 / K}$
$C =\frac{ Q _0 K}{ V _0}$
$C = C _0 K$
or $C = K \left(\frac{\epsilon_0 A}{d}\right)$
Therefore, the value of capacitance C of a parallel plate capacitor completely filled with dielectric material is K times the value of air or vacuum capacitance $C_0$.
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