Parallel plate capacitor... separation 5 mm... mica sheet 2 mm... draws 25\% more charge. Dielectric constant is ___.
Show Hint
- 2.0
- 1.0
- 1.5
- 2.5
The Correct Option is A
Solution and Explanation
A parallel plate capacitor initially has no dielectric material between its plates. When a mica sheet of thickness 2 mm is inserted (the total separation between the plates is 5 mm), it is stated that the capacitor draws 25% more charge. We need to find the dielectric constant of mica.
The capacitance of a capacitor without a dielectric is given by:
\(C_0 = \frac{\varepsilon_0 A}{d}\)
where:
- \(\varepsilon_0\) is the permittivity of free space,
- \(A\) is the area of the plates, and
- \(d\) is the separation between the plates.
With the dielectric (mica) inserted, the capacitor system now consists of two dielectric mediums: air and mica. The mica fills 2 mm of the space, while the air fills the remaining 3 mm. The effective capacitance \(C\) is given by:
\(C = \frac{\varepsilon_0 A}{d_{air}} + \frac{\varepsilon_ A}{d_{mica}}\)
where
- \(d_{air} = 3 \, \text{mm}\),
- \(d_{mica} = 2 \, \text{mm}\), and
- \(\varepsilon\) is the permittivity of the mica, which can be expressed as \(\varepsilon = K \cdot \varepsilon_0\), where \(K\) is the dielectric constant of mica.
The new capacitance becomes:
\(C = \varepsilon_0 A \left(\frac{1}{d_{air}} + \frac{K}{d_{mica}}\right)\).
We know the charge increases by 25%, meaning
\(C = 1.25 C_0\).
Substituting the values, we have:
\(\varepsilon_0 A \left(\frac{1}{3 \times 10^{-3}} + \frac{K}{2 \times 10^{-3}}\right) = 1.25 \times \frac{\varepsilon_0 A}{5 \times 10^{-3}}\).
Simplifying, we find:
\(\left(\frac{1}{3 \times 10^{-3}} + \frac{K}{2 \times 10^{-3}}\right) = \frac{1.25}{5 \times 10^{-3}}\).
Solving for \(K\),
\(\frac{1}{3} + \frac{K}{2} = \frac{1.25}{5} = 0.25\).
\(K = \frac{5}{2} - \frac{2}{3} = \frac{10 - 2}{3} = 2\).
Therefore, the dielectric constant of mica is \(2.0\).
Hence, the correct answer is 2.0.
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