The capacitance of an air-filled parallel-plate capacitor is 60 pF. When a dielectric slab whose thickness is half the distance between the plates, is placed on one of the plates covering it entirely, the capacitance becomes 86 pF. Neglecting the fringing effects, the relative permittivity of the dielectric is __________ (up to 2 decimal places).
We know that,
Capacitance if a parallel plate capacitor is,
\(C = \frac{{{\epsilon_0}A}}{d}\)
\({C_1} = \frac{{{\epsilon_0}{A_1}}}{{{d_1}}} = 60\;pF\)
When a dielectric slab whose thickness is half the distance between placed on one of the plates covering, the capacitance becomes = 86 pF
C2 = 86 pF
\({C_2} = \frac{{{\epsilon_0}{A_1}}}{{\frac{{{d_1}}}{2}}},\frac{{{\epsilon_0}{\epsilon_r}{A_1}}}{{\frac{{{d_1}}}{2}}}\)
\(= \frac{{2{\epsilon_0}{A_1}}}{{{d_1}}},\frac{{2{\epsilon_0}{A_1}{\epsilon_r}}}{{{d_1}}}\)
= 120, 120 ϵr
\({C_2} = \frac{{120 \times 120 \times {\epsilon_r}}}{{120 + 120{\epsilon_r}}}\)
\(\Rightarrow 87 = \frac{{120 \times 120{\epsilon_r}}}{{120\left( {1 + {\epsilon_r}} \right)}}\)
⇒ 86 + 86 ϵr = 120 ϵr
\(\Rightarrow {\epsilon_r} = \frac{{86}}{{34}} = 2.52\)
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