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Capacitor question

2021 · 18 Mar · Shift 1 · Q67
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  5. /2021 · 18 Mar · Shift 1 · Q67

Capacitor question

2021 · 18 Mar · Shift 1 · Q67

JEE MainPhysicsCapacitorNumerical+4 / −1
A parallel plate capacitor has plate area 100 m2 and plate separation of 10 m. The space between the plates is filled up to a thickness 5 m with a material of dielectric constant of 10. The resultant capacitance of the system is 'x' pF. The value of ε\varepsilonε 0 = 8.85 ×\times× 10 −-− 12 F.m −-− 1. The value of 'x' to the nearest integer is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 161

  1. Interpret the geometry

A parallel plate capacitor has:

  • Plate area: A=100 m2A = 100\,\text{m}^2A=100m2
  • Plate separation: d=10 md = 10\,\text{m}d=10m

The space between plates is filled up to thickness 5 m5\,\text{m}5m with dielectric constant K=10K = 10K=10. So the remaining 5 m5\,\text{m}5m is air/vacuum.

Since the dielectric fills the gap along the separation direction, the system is equivalent to two capacitors in series:

  • one of thickness d1=5 md_1 = 5\,\text{m}d1​=5m and permittivity ε1=Kε0=10ε0\varepsilon_1 = K\varepsilon_0 = 10\varepsilon_0ε1​=Kε0​=10ε0​
  • one of thickness d2=5 md_2 = 5\,\text{m}d2​=5m and permittivity ε2=ε0\varepsilon_2 = \varepsilon_0ε2​=ε0​
  1. Capacitance of each part

For a parallel plate capacitor,

C=εAdC = \frac{\varepsilon A}{d}C=dεA​

So,

C1=10ε0A5C_1 = \frac{10\varepsilon_0 A}{5}C1​=510ε0​A​ C2=ε0A5C_2 = \frac{\varepsilon_0 A}{5}C2​=5ε0​A​

These are in series, hence

1Ceq=1C1+1C2\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2}Ceq​1​=C1​1​+C2​1​

But it is simpler to use the layered dielectric formula:

Ceq=ε0Ad1K+d2C_{\text{eq}} = \frac{\varepsilon_0 A}{\frac{d_1}{K} + d_2}Ceq​=Kd1​​+d2​ε0​A​

Substitute values:

Ceq=8.85×10−12×100510+5C_{\text{eq}} = \frac{8.85\times 10^{-12}\times 100}{\frac{5}{10} + 5}Ceq​=105​+58.85×10−12×100​ =8.85×10−100.5+5= \frac{8.85\times 10^{-10}}{0.5 + 5}=0.5+58.85×10−10​ =8.85×10−105.5= \frac{8.85\times 10^{-10}}{5.5}=5.58.85×10−10​ =1.609×10−10 F= 1.609\times 10^{-10}\,\text{F}=1.609×10−10F
  1. Convert into pF

Since

1 pF=10−12 F1\,\text{pF} = 10^{-12}\,\text{F}1pF=10−12F

Therefore,

x=1.609×10−1010−12=160.9 pFx = \frac{1.609\times 10^{-10}}{10^{-12}} = 160.9\,\text{pF}x=10−121.609×10−10​=160.9pF

To the nearest integer,

x=161x = 161x=161
  1. Comparison with stored answer

Derived answer: 161161161

Stored correct answer: 161161161

They match.

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