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A parallel plate capacitor with air between the plates has a capacitance of \[8\mu F\] (\[1\mu F \times {10^{ - 12}}\]). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

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Last updated date: 20th Jun 2024
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Answer
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Hint The capacitance of a capacitor is directly proportional to the dielectric constant and inversely proportional to the distance between the plates. Compare the two capacitances to each other by division.
Formula used: In this solution we will be using the following formulae;
\[C = \dfrac{{K\varepsilon A}}{d}\] where \[C\] is the capacitance of a capacitor, \[K\] is the dielectric constant of the material between the capacitor plates, \[\varepsilon \] is the permittivity of free space, \[A\] is the area of the capacitor plates, and \[d\] is the distance between the plates.

Complete Step-by-Step solution:
According to the question, the first air is between the plates, hence, the dielectric constant is 1. Generally, the capacitance of a capacitor is given by
\[C = \dfrac{{K\varepsilon A}}{d}\] where\[K\] is the dielectric constant of the material between the capacitor plates, \[\varepsilon \] is the permittivity of free space, \[A\] is the area of the capacitor plates, and \[d\] is the distance between the plates.
Hence, initially,
\[C = \dfrac{{\varepsilon A}}{d} = 8 \times {10^{ - 12}}\]
Then, capacitance after a dielectric material is added, and the distance between the plate is halved will be given as
\[{C_2} = \dfrac{{2K\varepsilon A}}{d}\]
Then, we compare the two capacitances by dividing, we have
\[\dfrac{{{C_2}}}{C} = \dfrac{{2K\varepsilon A}}{d} \div \dfrac{{\varepsilon A}}{d}\]
\[ \Rightarrow \dfrac{{{C_2}}}{C} = \dfrac{{2K\varepsilon A}}{d} \times \dfrac{d}{{\varepsilon A}}\]
Which by cancellation will give the expression
\[\dfrac{{{C_2}}}{C} = 2K\]
\[ \Rightarrow {C_2} = 2KC\]
By inserting all known values, we have
\[{C_2} = 2 \times 6 \times 8 \times {10^{ - 12}} = 9.6 \times {10^{ - 11}}F\]

Note: Noting that in the final expression the area, and permittivity of free space where absent, this implies that we can find the expression without their knowledge. Hence, alternatively, from the knowledge that the capacitance is proportional to dielectric constant but inversely to distance, we can just write that generally,
\[C = k\dfrac{K}{d}\] where \[k\] is an arbitrary constant. Hence,
\[\dfrac{{{C_2}}}{C} = k\dfrac{K}{{\dfrac{d}{2}}} \div k\dfrac{1}{d} = k\dfrac{{2K}}{d} \times \dfrac{d}{k}\]
\[ \Rightarrow {C_2} = 2KC\]

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A parallel plate capacitor with air between the plates has a capacitance of \[8\mu F\] (\[1\mu F \times {10^{ - 12}}\]). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

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Electrostatic Potential and Capacitance Class 12 Physics - NCERT EXERCISE 2.5 | Vishal Kumar Sir
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