MHT CET202310 May 2023Evening ShiftPhysicsCapacitanceActual
Two dielectric slabs having dielectric constant ' K ₁ ' and ' K ₂ ' of thickness ' d 4 and 3 ~d 4 are inserted between the plates as shown in figure. The net capacitance between A and B is [ ₀ . is permittivity of free space]
Options
- A2 ~A ₀ ~d [ K ₁ ~K ₂ 3 ~K ₁+ K ₂ ]
- B3 ~A ₀ ~d [ K ₁+ K ₂ ~K ₁ ~K ₂ ]
- C3 A ₀ 2 ~d [ K ₁+ K ₂ ~K ₁ ~K ₂ ]
- D4 A ₀ d [ K₁ K₂ 3 K₁+K₂ ]
Correct answer
D. 4 A ₀ d [ K₁ K₂ 3 K₁+K₂ ]
Step-by-step solution
Capacity of 1^ st Capacitor, C₁= K₁ ₀ A d / 4 = 4 K₁ ₀ A d Capacity of 2^ nd Capacitor, C ₂= K ₂ ₀ ~A 3 ~d / 4 = 4 ~K ₂ ₀ ~A 3 ~d Equivalent capacitance 1 C = 1 C ₁ + 1 C ₂ aligned 1 C ₁ & = d 4 ~K ₁ ₀ ~A ; 1 C ₂ = 3 ~d 4 ~K ₂ ₀ ~A 1 C & = d 4 ~K ₁ ₀ ~A + 3 ~d 4 ~K ₂ ₀ ~A 1 C & = d 4 ₀ ~A [ 1 ~K ₁ + 3 ~K ₂ ] 1 C & = d 4 ₀ ~A [ K ₂+3 ~K ₁ ~K ₁ ~K ₂ ] C & = 4 ₀ ~A ~d [ K ₁ ~K ₂ 3 ~K ₁+ K ₂ ] aligned