A dielectric of thickness $5$ cm and a dielectric constant $10$ is introduced between the plates of a parallel plate capacitor having plate area $500 s q .$ cm and separation between the plates $10 c m$ . The capacitance of the capacitor with the dielectric slab is $ε_{0} = 8.8 \times 10^{- 12} C^{2} / N - m^{2}$

4.4 p F

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6.2 p F

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8 p F

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10 p F

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Solution

The correct option is C $8 p F$
We know
$C = \frac{k l_{0} A}{d}$
effective capacitors is series combination of $\frac{k l_{0} A}{5 c m} a n d \frac{l_{0} A}{5 c m}$
$∴ \frac{1}{C_{e f f}} = \frac{5 c m}{k l_{0} A} + \frac{5 c m}{k l_{0} A}$

$\Rightarrow C_{e f f} = \frac{k l_{0} A}{(5 + k 5) \times 10^{- 2}}$

$\Rightarrow C_{e f f} = \frac{10 \times 8.8 \times 10^{- 12} \times 500 \times 10^{- 4}}{55 \times 10^{- 2}}$
$\Rightarrow C_{e f f} = 8 p F$

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A dielectric of thickness 5 cm and a dielectric constant 10 is introduced between the plates of a parallel plate capacitor having plate area 500sq. cm and separation between the plates 10cm. The capacitance of the capacitor with the dielectric slab is ε0=8.8×10−12C2/N−m2

The correct option is C 8pFWe knowC=kl0Adeffective capacitors is series combination of kl0A5 cm and l0A5 cm∴1Ceff=5 cmkl0A+5 cmkl0A⇒Ceff=kl0A(5+k5)×10−2⇒Ceff=10×8.8×10−12×500×10−455×10−2⇒Ceff=8 pF