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Capacitor question

2022 · 24 Jun · Shift 1 · Q59
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  5. /2022 · 24 Jun · Shift 1 · Q59

Capacitor question

2022 · 24 Jun · Shift 1 · Q59

JEE MainPhysicsCapacitorMCQ+4 / −1
A parallel plate capacitor is formed by two plates each of area 30 π\piπ cm2 separated by 1 mm. A material of dielectric strength 3.6 ×\times× 107 Vm −-− 1 is filled between the plates. If the maximum charge that can be stored on the capacitor without causing any dielectric breakdown is 7 ×\times× 10 −-− 6C, the value of dielectric constant of the material is : [Use 14πε0=9×109{1 \over {4\pi {\varepsilon _0}}} = 9 \times {10^9}4πε0​1​=9×109 Nm2 C −-− 2]
  1. A
    1.66
  2. B
    1.75
  3. C
    2.25
  4. D
    2.33
View written solutionFree

Correct answer: D

  1. Given data
  • Plate area: A=30π cm2=30π×10−4 m2=3π×10−3 m2A = 30\pi\ \text{cm}^2 = 30\pi \times 10^{-4}\ \text{m}^2 = 3\pi \times 10^{-3}\ \text{m}^2A=30π cm2=30π×10−4 m2=3π×10−3 m2
  • Separation between plates: d=1 mm=10−3 md = 1\ \text{mm} = 10^{-3}\ \text{m}d=1 mm=10−3 m
  • Dielectric strength: Emax⁡=3.6×107 V/mE_{\max} = 3.6 \times 10^7\ \text{V/m}Emax​=3.6×107 V/m
  • Maximum charge stored: Qmax⁡=7×10−6 CQ_{\max} = 7 \times 10^{-6}\ \text{C}Qmax​=7×10−6 C

We need to find dielectric constant KKK.

  1. Condition for dielectric breakdown

For a capacitor filled with dielectric, Vmax⁡=Emax⁡dV_{\max} = E_{\max} dVmax​=Emax​d

So, Vmax⁡=3.6×107×10−3=3.6×104 VV_{\max} = 3.6 \times 10^7 \times 10^{-3} = 3.6 \times 10^4\ \text{V}Vmax​=3.6×107×10−3=3.6×104 V

  1. Use relation Q=CVQ = CVQ=CV

At maximum charge, Qmax⁡=CVmax⁡Q_{\max} = C V_{\max}Qmax​=CVmax​

Hence, C=Qmax⁡Vmax⁡=7×10−63.6×104C = \frac{Q_{\max}}{V_{\max}} = \frac{7 \times 10^{-6}}{3.6 \times 10^4}C=Vmax​Qmax​​=3.6×1047×10−6​

C=73.6×10−10C = \frac{7}{3.6} \times 10^{-10}C=3.67​×10−10

  1. Capacitance of parallel plate capacitor with dielectric

C=Kε0AdC = \frac{K\varepsilon_0 A}{d}C=dKε0​A​

Thus, K=Cdε0AK = \frac{Cd}{\varepsilon_0 A}K=ε0​ACd​

We are given: 14πε0=9×109\frac{1}{4\pi\varepsilon_0} = 9 \times 10^94πε0​1​=9×109

Therefore, ε0=14π×9×109=136π×109\varepsilon_0 = \frac{1}{4\pi \times 9 \times 10^9} = \frac{1}{36\pi \times 10^9}ε0​=4π×9×1091​=36π×1091​

Now, K=(73.6×10−10)(10−3)(136π×109)(3π×10−3)K = \frac{\left(\frac{7}{3.6} \times 10^{-10}\right)(10^{-3})}{\left(\frac{1}{36\pi \times 10^9}\right)(3\pi \times 10^{-3})}K=(36π×1091​)(3π×10−3)(3.67​×10−10)(10−3)​

Simplify denominator: (136π×109)(3π×10−3)=3π×10−336π×109=112×10−12\left(\frac{1}{36\pi \times 10^9}\right)(3\pi \times 10^{-3}) = \frac{3\pi \times 10^{-3}}{36\pi \times 10^9} = \frac{1}{12} \times 10^{-12}(36π×1091​)(3π×10−3)=36π×1093π×10−3​=121​×10−12

So, K=73.6×10−13112×10−12K = \frac{\frac{7}{3.6} \times 10^{-13}}{\frac{1}{12} \times 10^{-12}}K=121​×10−123.67​×10−13​

K=73.6×12×10−1K = \frac{7}{3.6} \times 12 \times 10^{-1}K=3.67​×12×10−1

K=843.6×10−1K = \frac{84}{3.6} \times 10^{-1}K=3.684​×10−1

K=23.33×10−1=2.33K = 23.33 \times 10^{-1} = 2.33K=23.33×10−1=2.33

  1. Match with options

The dielectric constant is 2.33\boxed{2.33}2.33​

So the correct option is D.

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