A parallel plate capacitor has plate area $100 c m^{2}$ and separation between the plates is 1 cm. A glass plat $(k_{1} = 6)$ of thickness 6 mm and an ebonite plate $(k_{2} = 4)$ of thickness 4 mm are inserted. Find C:

4.085 pF

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40.85 pF

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.4085 nF

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40.85 nF

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Solution

The correct option is B 40.85 pF
Capacitors are in series so equivalent capacitance is given by:
$C_{e q} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} = \frac{ϵ_{0} A k_{1} k_{2}}{k_{1} d_{2} + k_{2} d_{1}} = \frac{8.85 \times 10^{- 12} \times 10^{- 2} \times 6 \times 4}{(6 \times 6 + 4 \times 4) \times 10^{- 3}} = 4.085 \times 10^{- 11} F$

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A parallel plate capacitor has plate area 100cm2 and separation between the plates is 1 cm. A glass plat (k1=6) of thickness 6 mm and an ebonite plate (k2=4) of thickness 4 mm are inserted. Find C:

The correct option is B 40.85 pFCapacitors are in series so equivalent capacitance is given by: Ceq=C1C2C1+C2=ϵ0Ak1k2k1d2+k2d1=8.85×10−12×10−2×6×4(6×6+4×4)×10−3=4.085×10−11F

A parallel plate capacitor has plate area $100 c m^{2}$ and separation between the plates is 1 cm. A glass plat $(k_{1} = 6)$ of thickness 6 mm and an ebonite plate $(k_{2} = 4)$ of thickness 4 mm are inserted. Find C: