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Capacitor question

2025 · 28 Jan · Shift 2 · Q70
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Capacitor question

2025 · 28 Jan · Shift 2 · Q70

JEE MainPhysicsCapacitorMCQ+4 / −1
A parallel plate capacitor of capacitance 1 µF is charged to a potential difference of 20 V. The distance between plates is 1 µm. The energy density between plates of capacitor is :
  1. A
    1.8×1031.8 \times 10^31.8×103 J/m3
  2. B
    2×1022 \times 10^22×102 J/m3
  3. C
    2×10−42 \times 10^{-4}2×10−4 J/m3
  4. D
    1.8×1051.8 \times 10^51.8×105 J/m3
View written solutionFree

Correct answer: A

  1. Given data
  • Capacitance: C=1 μF=1×10−6 FC = 1\,\mu\text{F} = 1 \times 10^{-6}\,\text{F}C=1μF=1×10−6F
  • Potential difference: V=20 VV = 20\,\text{V}V=20V
  • Distance between plates: d=1 μm=1×10−6 md = 1\,\mu\text{m} = 1 \times 10^{-6}\,\text{m}d=1μm=1×10−6m

We need the energy density between the plates.

  1. Formula for energy density in electric field

Energy density is

u=12ε0E2u = \frac{1}{2}\varepsilon_0 E^2u=21​ε0​E2

For a parallel plate capacitor,

E=VdE = \frac{V}{d}E=dV​

So,

E=201×10−6=2×107 V/mE = \frac{20}{1 \times 10^{-6}} = 2 \times 10^7\,\text{V/m}E=1×10−620​=2×107V/m

  1. Substitute into energy density formula

u=12(8.85×10−12)(2×107)2u = \frac{1}{2}(8.85 \times 10^{-12})(2 \times 10^7)^2u=21​(8.85×10−12)(2×107)2

Now,

(2×107)2=4×1014(2 \times 10^7)^2 = 4 \times 10^{14}(2×107)2=4×1014

Hence,

u=12(8.85×10−12)(4×1014)u = \frac{1}{2}(8.85 \times 10^{-12})(4 \times 10^{14})u=21​(8.85×10−12)(4×1014)

u=12(35.4×102)u = \frac{1}{2}(35.4 \times 10^2)u=21​(35.4×102)

u=17.7×102u = 17.7 \times 10^2u=17.7×102

u=1.77×103 J/m3u = 1.77 \times 10^3\,\text{J/m}^3u=1.77×103J/m3

  1. Match with options

u≈1.8×103 J/m3u \approx 1.8 \times 10^3\,\text{J/m}^3u≈1.8×103J/m3

So the correct option is A.

  1. Verification with stored answer

Stored correct answer: A

My derived answer: A

They match.

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