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\(Nm^2C^{-2}\)]
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 \( 1 \, \text{cm}^2 = 10^{-4} \, \text{m}^2 \)).
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}}
