A parallel plate capacitor has plate area 100 $m^{2}$ and plate separation of 10m. The space between the plates is filled upto a thickness 5 m with a material of dielectric constant of 10. The resultant capacitance of the system is ‘x’ pF .The value of $ε_{0} = 8.85 \times 10^{- 12} F . m^{- 1}$
The value of ‘x’ to the nearest integer is __

Question

Infinity Learn · Accepted Answer

$$C=$$ε$$0Ad$$−$$t+t/k$$   ⇒$$C=8.85$$×$$10$$−$$12$$×$$10010$$−$$5+510=160pF$$

Insertion of dielectric in capacitor

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$C = \frac{ε_{0} A}{d - t + t / k} \Rightarrow C = \frac{8.85 \times 10^{- 12} \times 100}{10 - 5 + \frac{5}{10}} = 160 p F$

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Insertion of dielectric in capacitor

Remember concepts with our Masterclasses.

C=ε0Ad−t+t/k ⇒C=8.85×10−12×10010−5+510=160pF

Ready to Test Your Skills?

free study material

$C = \frac{ε_{0} A}{d - t + t / k} \Rightarrow C = \frac{8.85 \times 10^{- 12} \times 100}{10 - 5 + \frac{5}{10}} = 160 p F$