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Capacitor question

2014 · Shift 0 · Q59
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Capacitor question

2014 · Shift 0 · Q59

JEE MainPhysicsCapacitorMCQ+4 / −1
A parallel plate capacitor is made of two circular plates separated by a distance 5mm5mm5mm and with a dielectric of dielectric constant 2.22.22.2 between them. When the electric field in the dielectric is 3×104 V/m3 \times {10^4}\,V/m3×104V/m the charge density of the positive plate will be close to:
  1. A
    6×10−7  C/m26 \times {10^{ - 7}}\,\,C/{m^2}6×10−7C/m2
  2. B
    3×10−7  C/m23 \times {10^{ - 7}}\,\,C/{m^2}3×10−7C/m2
  3. C
    3×104  C/m23 \times {10^4}\,\,C/{m^2}3×104C/m2
  4. D
    6×104  C/m26 \times {10^4}\,\,C/{m^2}6×104C/m2
View written solutionFree

Correct answer: A

  1. Use the relation between electric field and surface charge density in a dielectric

For a parallel plate capacitor filled with a dielectric of relative permittivity KKK, the electric field is

E=σKε0E=\frac{\sigma}{K\varepsilon_0}E=Kε0​σ​

So,

σ=Kε0E\sigma = K\varepsilon_0 Eσ=Kε0​E

where:

  • K=2.2K=2.2K=2.2
  • ε0=8.85×10−12 F/m\varepsilon_0 = 8.85\times 10^{-12}\,\text{F/m}ε0​=8.85×10−12F/m
  • E=3×104 V/mE=3\times 10^4\,\text{V/m}E=3×104V/m
  1. Substitute the values

σ=2.2×8.85×10−12×3×104\sigma = 2.2 \times 8.85\times 10^{-12} \times 3\times 10^4σ=2.2×8.85×10−12×3×104

First multiply the numerical factors:

2.2×8.85=19.472.2\times 8.85 = 19.472.2×8.85=19.47

19.47×3=58.4119.47 \times 3 = 58.4119.47×3=58.41

Now combine powers of 10:

10−12×104=10−810^{-12}\times 10^4 = 10^{-8}10−12×104=10−8

Thus,

σ=58.41×10−8 C/m2\sigma = 58.41\times 10^{-8}\,\text{C/m}^2σ=58.41×10−8C/m2

σ=5.841×10−7 C/m2\sigma = 5.841\times 10^{-7}\,\text{C/m}^2σ=5.841×10−7C/m2

  1. Choose the closest option

σ≈6×10−7 C/m2\sigma \approx 6\times 10^{-7}\,\text{C/m}^2σ≈6×10−7C/m2

So the correct option is A.

  1. Note

The plate separation 5 mm5\,\text{mm}5mm is not needed here, because the electric field is already given directly.

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