A parallel plate capacitor is formed by two plates each of area 30p cm2 separated by 1 mm. A material of dielectric strength 3.6 × 107 Vm–1 is filled between the plates. If the maximum charge that can be stored on the capacitor without causing any dielectric breakdown is 7 × 10–6 C, the value of dielectric constant of the material is : [\(\frac{1}{4πε0}\)=9×109\(Nm^2C^{-2}\)]
- 1.66
- 1.75
- 2.25
- 2.33
The Correct Option is D
Solution and Explanation
To solve this problem, we need to find the dielectric constant of the material between the plates of the capacitor. Let's go through the problem step-by-step:
- The formula for the capacitance \( C \) of a parallel plate capacitor is given by:
C = \frac{{\varepsilon_0 \varepsilon_r A}}{{d}}
where:
- \( \varepsilon_0 \) is the permittivity of free space.
- \( \varepsilon_r \) is the relative permittivity (dielectric constant) of the material.
- \( A \) is the area of the plates.
- \( d \) is the distance between the plates.
- Using the values provided:
- Area \( A = 30\pi \times 10^{-4} \, \text{m}^2 \) (since \( 1 \, \text{cm}^2 = 10^{-4} \, \text{m}^2 \)).
- Distance \( d = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \).
- The capacitor can store a maximum charge \( Q \) without breakdown: Q = C \cdot V\) where \( V \) is the breakdown voltage given by: V = E \cdot d and \( E \) is the dielectric strength \( 3.6 \times 10^7 \, \text{Vm}^{-1} \).
- Substitute \( V \) into the formula for \( Q \): Q = \frac{{\varepsilon_0 \varepsilon_r A}}{{d}} \cdot E \cdot d
- Simplify to find \( \varepsilon_r \):
- Given charge \( Q = 7 \times 10^{-6} \, \text{C}\).
- Equate and rearrange: \varepsilon_0 \varepsilon_r A E = Q \implies \varepsilon_r = \frac{Q}{{\varepsilon_0 A E}}
- Substitute the known values:
- \(\varepsilon_0 = \frac{1}{4\pi \times 9 \times 10^9} \, \text{Fm}^{-1}\).
- \(A = 30\pi \times 10^{-4} \, \text{m}^2\).
- \(E = 3.6 \times 10^7 \, \text{Vm}^{-1}\).
Calculating the values: \varepsilon_r = \frac{Q \cdot 4\pi \cdot 9 \times 10^9}{30\pi \times 10^{-4} \times 3.6 \times 10^7}
This simplifies to: \varepsilon_r = 2.33
Therefore, the dielectric constant of the material is 2.33, which matches the correct answer.
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