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Standard XII

Physics

New Capacitance in Dielectric

The area of t...

Question

The area of the positive plate is $125 c m^{2}$ and the area of the negative plate is $100 c m^{2}$ , They are parallel to each other and are separated by 0.5 cm . The capacity of a condenser with air as dielectric is :
$(ε_{0} = 8.9 \times 10^{- 12} C^{2} N^{- 1} M^{- 2})$

22.25 p F

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20.02 p F

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17.8 μ F

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17.8 p F

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Solution

The correct option is B $17.8 p F$
when area of plate are different we consider the area which is smaller to calculate capacitor,
$∴ c = \frac{ϵ_{\circ} A}{d}$
$\Rightarrow c = \frac{8.9 \times 10^{- 12} \times 100 \times 10^{- 4}}{0.5 \times 10^{- 2}}$
$\Rightarrow c = 17.8 p F$

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The area of the positive plate is 125cm2 and the area of the negative plate is 100cm2, They are parallel to each other and are separated by 0.5 cm . The capacity of a condenser with air as dielectric is :(ε0=8.9×10−12C2N−1M−2)

The correct option is B 17.8pFwhen area of plate are different we consider the area which is smaller to calculate capacitor,∴c=ϵ∘ Ad⇒c=8.9×10−12×100×10−40.5×10−2⇒c=17.8 pF

The area of the positive plate is $125 c m^{2}$ and the area of the negative plate is $100 c m^{2}$ , They are parallel to each other and are separated by 0.5 cm . The capacity of a condenser with air as dielectric is :
$(ε_{0} = 8.9 \times 10^{- 12} C^{2} N^{- 1} M^{- 2})$

The correct option is B $17.8 p F$
when area of plate are different we consider the area which is smaller to calculate capacitor,
$∴ c = \frac{ϵ_{\circ} A}{d}$
$\Rightarrow c = \frac{8.9 \times 10^{- 12} \times 100 \times 10^{- 4}}{0.5 \times 10^{- 2}}$
$\Rightarrow c = 17.8 p F$