A parallel plate condenser has a capacitance \[50\mu F\] in air and \[100\mu F\] when immersed in an oil. The dielectric constant $k$ of the oil is
A. $0.45$
B. $0.55$
C. $1.10$
D. $2.20$
Answer
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Hint: To solve this question, we just need to put the values in the formula and little concept about capacitance and dielectric is enough. A capacitor is a device that stores electrical energy in the form of an electric field. It is made up of two conductors that are separated by a dielectric. The ability of a capacitor to store an electric charge is known as capacitance.
Formula used:
$C = k{\varepsilon _0}\dfrac{A}{d}$
Where, ${\varepsilon _0}$ is the permittivity of space, $k$ is the relative permittivity of dielectric material, $d$ is the separation between the plates and $A$ is the area of plates.
Complete step by step answer:
According to the question, capacitance of capacitor when in air is \[{C_1} = 50\mu F\] and capacitance of capacitor when in oil is \[{C_2} = 100\mu F\]. Now,
${C_1} = 50\mu F = \dfrac{{Ak{\varepsilon _0}}}{d}$
$\Rightarrow {C_1} = 50\mu F = \dfrac{{A{\varepsilon _0}}}{d}$....................(1)
Where $k = 1$ because the dielectric strength of air is $1$.
${C_2} = 110\mu F = \dfrac{{Ak{\varepsilon _0}}}{d}$
As we know that,capacitance of new capacitor \[100\mu F\] = $\dfrac{{Ak{\varepsilon _0}}}{d}$
After putting the value of $\dfrac{{A{\varepsilon _0}}}{d}$ from equation (1), we get;
$k = \dfrac{{110}}{{50}} \\
\therefore k = 2.20 $
Hence, the correct option is D.
Additional information: Dielectric material can be an insulator, because a dielectric has poor conductivity to electrical charge. It is also known as relative permittivity. The dielectric constant of a substance is defined as the ratio of the permittivity of the substance to the permittivity of the free space. It is denoted by the symbol kappa $k$ . It is expressed in units Farad per meter.
Note: It is to be noted that the capacitance of a capacitor can be increased if the dielectric constant is inserted between the plates. By increasing the capacitance, the ability of the capacitor to store opposite charges on the plates will increase. Dielectric constant is basically a number that has no dimensions. An insulator is used as a dielectric as it fills the space between the plates but if a metal is used then the field inside the metal will be zero.Therefore, there will be no charge stored.
Formula used:
$C = k{\varepsilon _0}\dfrac{A}{d}$
Where, ${\varepsilon _0}$ is the permittivity of space, $k$ is the relative permittivity of dielectric material, $d$ is the separation between the plates and $A$ is the area of plates.
Complete step by step answer:
According to the question, capacitance of capacitor when in air is \[{C_1} = 50\mu F\] and capacitance of capacitor when in oil is \[{C_2} = 100\mu F\]. Now,
${C_1} = 50\mu F = \dfrac{{Ak{\varepsilon _0}}}{d}$
$\Rightarrow {C_1} = 50\mu F = \dfrac{{A{\varepsilon _0}}}{d}$....................(1)
Where $k = 1$ because the dielectric strength of air is $1$.
${C_2} = 110\mu F = \dfrac{{Ak{\varepsilon _0}}}{d}$
As we know that,capacitance of new capacitor \[100\mu F\] = $\dfrac{{Ak{\varepsilon _0}}}{d}$
After putting the value of $\dfrac{{A{\varepsilon _0}}}{d}$ from equation (1), we get;
$k = \dfrac{{110}}{{50}} \\
\therefore k = 2.20 $
Hence, the correct option is D.
Additional information: Dielectric material can be an insulator, because a dielectric has poor conductivity to electrical charge. It is also known as relative permittivity. The dielectric constant of a substance is defined as the ratio of the permittivity of the substance to the permittivity of the free space. It is denoted by the symbol kappa $k$ . It is expressed in units Farad per meter.
Note: It is to be noted that the capacitance of a capacitor can be increased if the dielectric constant is inserted between the plates. By increasing the capacitance, the ability of the capacitor to store opposite charges on the plates will increase. Dielectric constant is basically a number that has no dimensions. An insulator is used as a dielectric as it fills the space between the plates but if a metal is used then the field inside the metal will be zero.Therefore, there will be no charge stored.
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