When a dielectric slab of thickness 6cm is introduced between the plates of parallel plate condenser, it is found that the distance between the plates has to be increased by 4cm to restore capacity to original value. The dielectric constant of the slab is:
$\begin{align}
& (A)\dfrac{3}{2} \\
& (B)\dfrac{2}{3} \\
& (C)3 \\
& (D)4 \\
\end{align}$
Answer
551.4k+ views
Hin:: In the first case, the capacitance without a dielectric will be calculated. And, in the second case the resultant capacitance due to the series combination of dielectric and air will be calculated. Since, in the second case after increasing the plate distance by 4cm, original capacitance is restored. Therefore, now we can equate these two capacitances to calculate the dielectric constant of the slab.
Complete answer:
Let the distance between the two plates of the capacitor be ‘d’. The area of the capacitor plates be ‘A’ and the dielectric constant of the capacitor be K. Then, in the first case the capacitance will be equal to:
$\Rightarrow {{C}_{1}}=\dfrac{A{{\varepsilon }_{0}}}{d}$ [Let this expression be equation number (1)]
Now, in the second case the capacitors are in series. These capacitances are due to air and dielectric slab.
The capacitance due to slab is equal to:
$\Rightarrow {{C}_{S}}=\dfrac{KA{{\varepsilon }_{0}}}{6}$
And, the capacitance due to air is:
$\Rightarrow {{C}_{A}}=\dfrac{A{{\varepsilon }_{0}}}{(d+4)-6}$
Hence, their equivalent capacitance can be calculated as:
$\Rightarrow \dfrac{1}{{{C}_{2}}}=\dfrac{1}{{{C}_{S}}}+\dfrac{1}{{{C}_{A}}}$
Since, the capacitance is restored to its original value,
$\therefore {{C}_{2}}={{C}_{1}}$
Using this in the above equation and putting the values of all the terms, we get:
$\begin{align}
& \Rightarrow \dfrac{d}{A{{\varepsilon }_{0}}}=\dfrac{6}{KA{{\varepsilon }_{0}}}+\dfrac{(d+4)-6}{A{{\varepsilon }_{0}}} \\
& \Rightarrow d=\dfrac{6}{K}+d-2 \\
& \Rightarrow \dfrac{6}{K}=2 \\
& \therefore K=3 \\
\end{align}$
Hence, the dielectric constant of the slab is calculated to be 3.
Hence, option (C) is the correct option.
Note:
In the last step of our calculation. Our equation seemed to be containing two variables, that are ‘d’ and ‘K’. But once we solve it, one of them is eliminated. This means, we shouldn’t worry if the value to a certain parameter is not known to us, and just proceed with our solution.
Complete answer:
Let the distance between the two plates of the capacitor be ‘d’. The area of the capacitor plates be ‘A’ and the dielectric constant of the capacitor be K. Then, in the first case the capacitance will be equal to:
$\Rightarrow {{C}_{1}}=\dfrac{A{{\varepsilon }_{0}}}{d}$ [Let this expression be equation number (1)]
Now, in the second case the capacitors are in series. These capacitances are due to air and dielectric slab.
The capacitance due to slab is equal to:
$\Rightarrow {{C}_{S}}=\dfrac{KA{{\varepsilon }_{0}}}{6}$
And, the capacitance due to air is:
$\Rightarrow {{C}_{A}}=\dfrac{A{{\varepsilon }_{0}}}{(d+4)-6}$
Hence, their equivalent capacitance can be calculated as:
$\Rightarrow \dfrac{1}{{{C}_{2}}}=\dfrac{1}{{{C}_{S}}}+\dfrac{1}{{{C}_{A}}}$
Since, the capacitance is restored to its original value,
$\therefore {{C}_{2}}={{C}_{1}}$
Using this in the above equation and putting the values of all the terms, we get:
$\begin{align}
& \Rightarrow \dfrac{d}{A{{\varepsilon }_{0}}}=\dfrac{6}{KA{{\varepsilon }_{0}}}+\dfrac{(d+4)-6}{A{{\varepsilon }_{0}}} \\
& \Rightarrow d=\dfrac{6}{K}+d-2 \\
& \Rightarrow \dfrac{6}{K}=2 \\
& \therefore K=3 \\
\end{align}$
Hence, the dielectric constant of the slab is calculated to be 3.
Hence, option (C) is the correct option.
Note:
In the last step of our calculation. Our equation seemed to be containing two variables, that are ‘d’ and ‘K’. But once we solve it, one of them is eliminated. This means, we shouldn’t worry if the value to a certain parameter is not known to us, and just proceed with our solution.
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