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Capacitor question

2019 · 8 Apr · Shift 1 · Q71
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Capacitor question

2019 · 8 Apr · Shift 1 · Q71

JEE MainPhysicsCapacitorMCQ+4 / −1
Voltage rating of a parallel plate capacitor is 500V. Its dielectric can withstand a maximum electric field of 106 V/m. The plate area is 10–4 m2. What is the dielectric constant is the capacitance is 15 pF? (given ε\varepsilonε 0 = 8.86 × 10–12 C2/Nm2)
  1. A
    8.5
  2. B
    4.5
  3. C
    3.8
  4. D
    6.2
View written solutionFree

Correct answer: A

  1. Use voltage rating and dielectric strength to find plate separation

For a parallel plate capacitor, the maximum electric field is Emax⁡=Vmax⁡dE_{\max}=\frac{V_{\max}}{d}Emax​=dVmax​​ So, d=Vmax⁡Emax⁡=500106=5×10−4 md=\frac{V_{\max}}{E_{\max}}=\frac{500}{10^6}=5\times 10^{-4}\ \text{m}d=Emax​Vmax​​=106500​=5×10−4 m

  1. Use capacitance formula with dielectric

For a parallel plate capacitor filled with dielectric, C=Kε0AdC=K\varepsilon_0\frac{A}{d}C=Kε0​dA​ Hence, K=Cdε0AK=\frac{Cd}{\varepsilon_0 A}K=ε0​ACd​

Given:

  • C=15 pF=15×10−12 FC=15\ \text{pF}=15\times 10^{-12}\ \text{F}C=15 pF=15×10−12 F
  • d=5×10−4 md=5\times 10^{-4}\ \text{m}d=5×10−4 m
  • ε0=8.86×10−12\varepsilon_0=8.86\times 10^{-12}ε0​=8.86×10−12
  • A=10−4 m2A=10^{-4}\ \text{m}^2A=10−4 m2

Substitute: K=(15×10−12)(5×10−4)(8.86×10−12)(10−4)K=\frac{(15\times 10^{-12})(5\times 10^{-4})}{(8.86\times 10^{-12})(10^{-4})}K=(8.86×10−12)(10−4)(15×10−12)(5×10−4)​

  1. Simplify

K=75×10−168.86×10−16=758.86K=\frac{75\times 10^{-16}}{8.86\times 10^{-16}}=\frac{75}{8.86}K=8.86×10−1675×10−16​=8.8675​

K≈8.47K\approx 8.47K≈8.47

Thus, K≈8.5K\approx 8.5K≈8.5

  1. Check options
  • A: 8.58.58.5 ✅
  • B: 4.54.54.5 ❌
  • C: 3.83.83.8 ❌
  • D: 6.26.26.2 ❌

Therefore, the correct option is A.

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