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Physics Question 49 – JEE-MAIN 2025

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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 ϵo=8.85×10−12 F/m)

The two dielectric layers are in series, forming an equivalent capacitor, which is then connected in parallel with another capacitor.

Step 1: Calculate the equivalent capacitance of the parallel plate capacitor with dielectrics✦ Active

The two dielectric layers are in series. Each layer has thickness d/2. The capacitance of the first layer (k1=5) is C1=k1ϵoAd/2=2×5×ϵoAd=10ϵoAd. The capacitance of the second layer (k2=3) is C2=k2ϵoAd/2=2×3×ϵoAd=6ϵoAd. The equivalent capacitance Ceq1 for these two in series is:

1Ceq1=1C1+1C2=d10ϵoA+d6ϵoA=dϵoA(110+16)=dϵoA(3+530)=4d15ϵoA

Therefore, Ceq1=15ϵoA4d. Substitute the given values: A=4×10−4 m2, d=1.77×10−3 m, ϵo=8.85×10−12 F/m.

Ceq1=15×(8.85×10−12)×(4×10−4)4×(1.77×10−3)=15×8.85×10−161.77×10−3=15×5×10−13=75×10−13 F=7.5×10−12 F=7.5 pF
Step 2: Calculate the total effective capacitance○ Expand

The capacitor Ceq1 is connected in parallel with another capacitor of capacitance Cadditional=7.5 pF. For parallel combination, the total effective capacitance Ctotal is the sum of individual capacitances.

Ctotal=Ceq1+Cadditional=7.5 pF+7.5 pF=15.0 pF
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