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Capacitor question

2019 · 12 Jan · Shift 2 · Q54
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Capacitor question

2019 · 12 Jan · Shift 2 · Q54

JEE MainPhysicsCapacitorMCQ+4 / −1
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\varepsilonε 0 = 8.85 ×\times× 10 −-− 12 C2N−m2{{{C^2}} \over {N - {m^2}}}N−m2C2​)
  1. A
    9.85 ×\times× 10–10 C
  2. B
    8.85 ×\times× 10–10 C
  3. C
    6.85 ×\times× 10–10 C
  4. D
    7.85 × 10–10 C
View written solutionFree

Correct answer: B

  1. Use the electric field between parallel plates

For a parallel plate capacitor, the electric field between the plates is

E=σε0E = \frac{\sigma}{\varepsilon_0}E=ε0​σ​

where σ\sigmaσ is the surface charge density.

Also,

σ=QA\sigma = \frac{Q}{A}σ=AQ​

So,

E=Qε0AE = \frac{Q}{\varepsilon_0 A}E=ε0​AQ​

Hence,

Q=ε0AEQ = \varepsilon_0 A EQ=ε0​AE

  1. Substitute the given values

Given:

  • A=1 m2A = 1\, \text{m}^2A=1m2
  • E=100 N/CE = 100\, \text{N/C}E=100N/C
  • ε0=8.85×10−12 C2/(N⋅m2)\varepsilon_0 = 8.85 \times 10^{-12}\, \text{C}^2/(\text{N}\cdot \text{m}^2)ε0​=8.85×10−12C2/(N⋅m2)

Therefore,

Q=(8.85×10−12)(1)(100)Q = (8.85 \times 10^{-12})(1)(100)Q=(8.85×10−12)(1)(100)

Q=8.85×10−10 CQ = 8.85 \times 10^{-10}\, \text{C}Q=8.85×10−10C

  1. Match with the options

Q=8.85×10−10 CQ = 8.85 \times 10^{-10}\, \text{C}Q=8.85×10−10C

So the correct option is:

B: 8.85×10−108.85 \times 10^{-10}8.85×10−10 C

  1. Note

The plate separation 0.1 m0.1\,\text{m}0.1m is not needed here, because charge is directly related to field by

Q=ε0AEQ = \varepsilon_0 A EQ=ε0​AE

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