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Capacitor question

2025 · 8 Apr · Shift 2 · Q73
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  5. /2025 · 8 Apr · Shift 2 · Q73

Capacitor question

2025 · 8 Apr · Shift 2 · Q73

JEE MainPhysicsCapacitorNumerical+4 / −1
JEE Main 2025 (Online) 8th April Evening Shift Physics - Capacitor Question 6 English Space between the plates of a parallel plate capacitor of plate area 4 cm2 and separation of 1.77 mm, is filled with uniform dielectric materials with dielectric constants (3 and 5) as shown in figure. Another capacitor of capacitance 7.5 pF is connected in parallel with it. The effective capacitance of this combination is _ pF. (Given ϵ0=8.85×10−12\epsilon_0 = 8.85 \times 10^{-12}ϵ0​=8.85×10−12 F/m)
Numerical answer
View written solutionFree

Correct answer: 15.5 PF, IF INTEGER ROUNDING IS INTENDED, 16

  1. Interpret the arrangement

For a parallel plate capacitor filled with two dielectrics side by side (as shown typically in such figures), each dielectric occupies half the plate area, so the system behaves like two capacitors in parallel.

Thus,

C=C1+C2C = C_1 + C_2C=C1​+C2​

with

C1=k1ϵ0A/2d,C2=k2ϵ0A/2dC_1 = \frac{k_1\epsilon_0 A/2}{d}, \qquad C_2 = \frac{k_2\epsilon_0 A/2}{d}C1​=dk1​ϵ0​A/2​,C2​=dk2​ϵ0​A/2​

So,

C=ϵ0A2d(k1+k2)C = \frac{\epsilon_0 A}{2d}(k_1+k_2)C=2dϵ0​A​(k1​+k2​)

Given:

  • A=4 cm2=4×10−4 m2A = 4\,\text{cm}^2 = 4\times 10^{-4}\,\text{m}^2A=4cm2=4×10−4m2
  • d=1.77 mm=1.77×10−3 md = 1.77\,\text{mm} = 1.77\times 10^{-3}\,\text{m}d=1.77mm=1.77×10−3m
  • k1=3k_1 = 3k1​=3, k2=5k_2 = 5k2​=5
  • ϵ0=8.85×10−12 F/m\epsilon_0 = 8.85\times 10^{-12}\,\text{F/m}ϵ0​=8.85×10−12F/m
  1. Compute the capacitance of the dielectric-filled capacitor
C=8.85×10−12×4×10−42×1.77×10−3(3+5)C = \frac{8.85\times 10^{-12} \times 4\times 10^{-4}}{2\times 1.77\times 10^{-3}}(3+5)C=2×1.77×10−38.85×10−12×4×10−4​(3+5)

Since (3+5)=8(3+5)=8(3+5)=8,

C=8.85×10−12×4×10−4×82×1.77×10−3C = \frac{8.85\times 10^{-12} \times 4\times 10^{-4} \times 8}{2\times 1.77\times 10^{-3}}C=2×1.77×10−38.85×10−12×4×10−4×8​

First simplify:

4×82=16\frac{4\times 8}{2} = 1624×8​=16

so,

C=16×8.85×10−161.77×10−3C = \frac{16\times 8.85\times 10^{-16}}{1.77\times 10^{-3}}C=1.77×10−316×8.85×10−16​

Now,

16×8.85=141.616\times 8.85 = 141.616×8.85=141.6

therefore,

C=141.6×10−161.77×10−3C = \frac{141.6\times 10^{-16}}{1.77\times 10^{-3}}C=1.77×10−3141.6×10−16​ C=(141.61.77)×10−13C = \left(\frac{141.6}{1.77}\right)\times 10^{-13}C=(1.77141.6​)×10−13 141.61.77=80\frac{141.6}{1.77} = 801.77141.6​=80

Hence,

C=80×10−13=8×10−12 F=8 pFC = 80\times 10^{-13} = 8\times 10^{-12}\,\text{F} = 8\,\text{pF}C=80×10−13=8×10−12F=8pF
  1. Add the 7.5 pF capacitor in parallel

For parallel combination,

Ceq=8+7.5=15.5 pFC_{\text{eq}} = 8 + 7.5 = 15.5\,\text{pF}Ceq​=8+7.5=15.5pF
  1. Final integer answer

Since this is an integer-type question, the effective capacitance is

16\boxed{16}16​

if rounded to the nearest integer.

  1. Compare with stored answer

Stored correct answer = 151515

My derived value is 15.5 pF15.5\,\text{pF}15.5pF, which rounds to 161616, not 151515. So I do not agree with the stored answer.

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