Verified Solution • 2025–26 Session moderate

QJEEM22Phe3fc: JEE Main 2022 Physics - Class 12 JEEM22Phe3fc Solution: A slab of dielectric constant has the same cross-sectional area …

A slab of dielectric constant K\mathrm{K} has the same cross-sectional area as the plates of a parallel plate capacitor and thickness 34 d\frac{3}{4} \mathrm{~d} , where d\mathrm{d} is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be :

(Given C0\mathrm{C}_{0} = capacitance of capacitor with air as medium between plates.)

Multiple Choice Options

  • A 4KC03+K\frac{4 K C_{0}}{3+K}
  • B 3KC03+K\frac{3 K C_{0}}{3+K}
  • C 3+K4KC0\frac{3+K}{4 K C_{0}}
  • D K4+K\frac{K}{4+K}

Step-by-Step Verified Solution

C0=ε0Ad{C_0} = {{{\varepsilon _0}A} \over d}

C=ε0Ad−3d4+3d4K=4ε0AK3d+KdC = {{{\varepsilon _0}A} \over {d - {{3d} \over 4} + {{3d} \over {4K}}}} = {{4{\varepsilon _0}AK} \over {3d + Kd}}

=4KC03+K= {{4K{C_0}} \over {3 + K}}

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