Physics Circle home Student Login

Home›Chapters›Capacitors PYQs›Dielectrics›Question 37

JEE Main 2025 (8 Apr, Shift 2) · Dielectrics

Dielectrics question from JEE Main 2025 (8 Apr, Shift 2)

Q37JEE Main 2025 · 8 APR · SHIFT 2Medium

Dielectrics

Figure

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} F/m)

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 15

  1. The two layers (d/2d/2 each) are in series: C5=5ϵ0Ad/2=10ϵ0AdC_5 = \frac{5\epsilon_0 A}{d/2} = 10\frac{\epsilon_0 A}{d}, C3=6ϵ0AdC_3 = 6\frac{\epsilon_0 A}{d}.
  2. C=10×616ϵ0Ad=3.75ϵ0AdC = \frac{10 \times 6}{16}\frac{\epsilon_0 A}{d} = 3.75\frac{\epsilon_0 A}{d}
  3. ϵ0Ad=8.85×10−12×4×10−41.77×10−3=2×10−12 F=2 pF\frac{\epsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 4 \times 10^{-4}}{1.77 \times 10^{-3}} = 2 \times 10^{-12}\text{ F} = 2\text{ pF}, so C=7.5 pFC = 7.5\text{ pF}.
  4. In parallel with 7.5 pF7.5\text{ pF}: Ceff=7.5+7.5=15 pFC_{eff} = 7.5 + 7.5 = 15\text{ pF}.
Report a mistake

More questions on Dielectrics