A parallel plate capacitor with air between the plates has a capacitance of 10 pF. The capacitance, if the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant 4 is:

80 pF

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96 pF

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100 pF

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120 pF

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Solution

The correct option is A 80 pF
Here
$K = 10 p F$ $K = 4$ $C_{1} = \frac{ε_{0} A}{d} = 10 μ F . . . (i)$
and $C_{2} = K \frac{ε_{0} A}{d / 2} = \frac{4 \times 2 ε_{0} A}{d} = 8 \times 10 = 80 p F$ (Using (i))

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A parallel plate capacitor with air between the plates has a capacitance of $9 p F$ . The separation between its plates is $d$ . The space between the plate is now partially filled with two dielectrics. One of the dielectric has dielectric constant $K_{1} = 3$ and thickness $\frac{d}{3}$ while the other one has dielectric constant $K_{2} = 6$ and thickness $\frac{d}{3}$ . The remaining $\frac{d}{3}$ space is filled with air. The new capacitance is :

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A parallel plate capacitor with air between the plates has a capacitance of 10 pF. The capacitance, if the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant 4 is:

The correct option is A 80 pFHereK=10pF K=4 C1=ε0Ad=10μF...(i)and C2=Kε0Ad/2=4×2ε0Ad=8×10=80pF (Using (i))

The correct option is A 80 pF
Here
$K = 10 p F$ $K = 4$ $C_{1} = \frac{ε_{0} A}{d} = 10 μ F . . . (i)$
and $C_{2} = K \frac{ε_{0} A}{d / 2} = \frac{4 \times 2 ε_{0} A}{d} = 8 \times 10 = 80 p F$ (Using (i))