Voltage rating of a parallel plate capacitor is 500 V. Its dielectric can withstand a maximum electric field of  10^{6}\; V/m. The plate area is 10^{-4}\;m^{2}. What is the dielectric constant if the capacitance is 15\; pF ?

(given \epsilon _{0}=8.86\times 10^{-12}C^{2}/Nm^{2})

  • Option 1)

    8.5

  • Option 2)

    6.2

  • Option 3)

    4.5

  • Option 4)

    3.8

 

Answers (1)

Area =10^{-4}m^{2}

C = 15\; PF

Voltage=500\; v

E=\frac{v}{d}\; \; \; \; \; (1)

C=\frac{k\varepsilon _{o}A}{d}\; \; \; \; \; \; \; (2)

from (1) & (2)

k=\frac{VC}{A\varepsilon _{o}E}=\frac{500\times 15\times 10^{12}}{10^{-4}\times 10^{6}\times 8.85\times 10^{-12}}

k\approx 8.5


Option 1)

8.5

Option 2)

6.2

Option 3)

4.5

Option 4)

3.8

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