The capacity of a parallel plate capacitor with no dielectric substance but with a separation of 0.4 cm is 2μF. The separation is reduced to half and it is filled with a dielectric substance of value 2.8. The final capacity of the capacitor is
11.2μF
19.2μF
22.4μF
15.6μF
A
11.2μF
B
22.4μF
C
19.2μF
D
15.6μF
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Solution
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Initial capacitance C=2uF=ε0Ad New capacitance C′=Kε0Ad′=2.8×ε0A(d/2)=5.6ε0Ad ⇒C′=5.6×2=11.2μF
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