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JEE Main 2018 (15 Apr, Shift 2) · Energy Stored & Force between Plates

Energy Stored & Force between Plates question from JEE Main 2018 (15 Apr, Shift 2)

Q30JEE Main 2018 · 15 APR · SHIFT 2Medium

Energy Stored & Force between Plates

A parallel plate capacitor with area 200 cm2 and separation between the plates 1.5 cm, is connected across a battery of emf V. If the force of attraction between the plates is 25×10−6 N,25 \times 10^{-6}\text{ N}, the value of V is approximately: (ϵ0=8.85×10−12 C2N⋅m2)\left(\epsilon_0 = 8.85 \times 10^{-12}\ \frac{C^2}{N \cdot m^2}\right)

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Answer: (a)

  1. F=ε0AV22d2⇒V2=2Fd2ε0AF = \frac{\varepsilon_0 A V^2}{2d^2} \Rightarrow V^2 = \frac{2Fd^2}{\varepsilon_0 A}
  2. A=2×10−2 m2A = 2 \times 10^{-2}\text{ m}^2, d=1.5×10−2 md = 1.5 \times 10^{-2}\text{ m}
  3. V2=2×25×10−6×2.25×10−48.85×10−12×2×10−2≈6.36×104V^2 = \frac{2 \times 25 \times 10^{-6} \times 2.25 \times 10^{-4}}{8.85 \times 10^{-12} \times 2 \times 10^{-2}} \approx 6.36 \times 10^{4}
  4. V≈252 VV \approx 252\text{ V}
  5. So the answer is (a) 250 V.
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