In a capacitor of capacitance 20 μ F, the distance between the plates is 2mm. If a dielectric slab of width 1mm and dielectric constant 2 is inserted between the plates, then the new capacitance is
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a
2 μ F
b
15.5 μ F
c
26.6 μ F
d
32 μ F
answer is C.
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Detailed Solution
C=ε0Ad and C′=ε0Ad−t+tK⇒CC′=d−t+tKd⇒20C′=2×10−3−1×10−3+1×10−322×10−3⇒C′=26.6μF
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In a capacitor of capacitance 20 μ F, the distance between the plates is 2mm. If a dielectric slab of width 1mm and dielectric constant 2 is inserted between the plates, then the new capacitance is