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JEE Main 2019 (12 Jan, Shift 2) · Parallel Plate Capacitor

Parallel Plate Capacitor question from JEE Main 2019 (12 Jan, Shift 2)

Q16JEE Main 2019 · 12 JAN · SHIFT 2Easy

Parallel Plate Capacitor

A parallel plate capacitor with plates of area 1 m2 each, are at a separation of 0.1 m. If the electric field between the plates is 100 N/C, the magnitude of charge on each plate is:

(Take ε\varepsilon0 = 8.85 ×\times 10−-12 C2N-m2\frac{C^2}{N\text{-}m^2})

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Answer: (b)

  1. E=σε0=Qε0A⇒Q=ε0AEE = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A} \Rightarrow Q = \varepsilon_0 AE
  2. Q=8.85×10−12×1×100=8.85×10−10 CQ = 8.85\times10^{-12}\times1\times100 = 8.85\times10^{-10}\text{ C}
  3. So the answer is (b) 8.85×10−10 C8.85\times10^{-10}\text{ C}.
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