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BITSAT 2009
Exam
BITSAT
Year
2009
Subject
Physics
Chapter
Capacitance

BITSAT 2009 · Physics · Capacitance

Calculate the area of the plates of a one farad parallel plate...

Calculate the area of the plates of a one farad parallel plate capacitor if separation between plates is 1mm1 \mathrm{ mm} and plates are in vacuum

Options

  1. A18×108m218 \times 10^{8} \mathrm{ m}^{2}
  2. B0.3×108m20.3 \times 10^{8} \mathrm{ m}^{2}
  3. C1.3×108m21.3 \times 10^{8} \mathrm{ m}^{2}
  4. D1.13×108m21.13 \times 10^{8} \mathrm{ m}^{2}

Correct answer

D. 1.13×108m21.13 \times 10^{8} \mathrm{ m}^{2}

Step-by-step solution

For a parallel plate capacitor C=ε0Ad\mathrm{C}=\frac{\varepsilon_{0} \mathrm{ A}}{\mathrm{ d}}∴A=Cdε0=1×10−38.85×10−12=1.13×108m2\begin{aligned} \therefore \quad \mathrm{A}=\frac{\mathrm{Cd}}{\varepsilon_{0}}=\frac{1 \times 10^{-3}}{8.85 \times 10^{-12}}\\ &=1.13 \times10^{8} \mathrm{ m}^{2} \end{aligned}$ThiscorrespondstoareaofsquareofsideThis corresponds to area of square of side10.6 \mathrm{ km}$ which shows that one farad is very large unit of capacitance.
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