
C1=21.77×10−35×4×10−4×8.85×10−12=20pFC2=21.77×10−33×4×10−4×8.85×10−12=12pFCeq=C1+C2C1C2=12+2012×20=7.5pF
Finally equivalent capacitance
(Ceq)final =7.5+7.5=15pF
JEE Main 2025 — Physics Electromagnetism

Space between the plates of a parallel plate capacitor of plate area 4 cm2 and separation of (d) 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 F/m)
Held on 8 Apr 2025 · Verified 6 Jul 2026.
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