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Question
A parallel-plate capacitor has plate area 100 cm2 and plate separation 1⋅0 cm. A glass plate (dielectric constant 6⋅0) of thickness 6⋅0 mm and an ebonite plate (dielectric constant 4⋅0) are inserted one over the other to fill the space between the plates of the capacitor. Find the new capacitance.
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Solution
The given system of the capacitor will behave as two capacitors connected in series.
Let the capacitances be C1 and C2.
Now,
`C_1 = (∈_0Ak_1)/d_1` and `C_2 = (∈_0Ak_2)/d_2`
Thus, the net capacitance is given by
`C = (C_1C_2)/(C_1+C_2)`
= `((∈_0Ak_1)/d_1 xx (∈_0Ak_2)/d_2)/((∈_0Ak_1)/d_1 +(∈_0Ak_2)/d_2)`
= `(∈_0A(k_1+k_2))/(k_1d_2+k_2d_1)`
= `((8.85 xx 10^-12) xx (10^-2) xx 24) / ((6 xx 4 xx 10^-3 + 4 xx 6 xx 10^-3)) = 4.425 xx 10^-11 C`
= 44.25 pF
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