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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,
C1=0Ak1d1 and C2=0Ak2d2
Thus, the net capacitance is given by
C=C1C2C1+C2 =0Ak1d1×0Ak2d20Ak1d1+0Ak2d2 =0Ak1+k2k1d2+k2d1 =(8.85×10-12)×(10-2)×24(6×4×10-3+4×6×10-3)=4.425×10-11C =44.25 pF

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