JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Capacitance

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A parallel plate capacitor with width 4 cm, length 8 cm and separation between the plates of 4 mm is connected to a battery of 20 V. A dielectric slab of dielectric constant 5 having length 1 cm, width 4 cm and thickness 4 mm is inserted between the plates of parallel plate capacitor. The electrostatic energy of this system will be ______ ε₀ J. (Where ε₀ is the permittivity of free space)

Answer: 240

Solution

The capacitor is partially filled with a dielectric. We treat it as two capacitors in parallel: one with dielectric (K=5, area 1×4 cm²) and one without (area 7×4 cm²).

C_{eff} = \frac{ε _{0} ( 7 \times 4 )}{4/10} + \frac{5 ε _{0} ( 1 \times 4 )}{4/10}

Converting to SI units (areas in m²):

C_{eff} = [\frac{28 ε _{0}}{0.4} + \frac{20 ε _{0}}{0.4}] \times 1 0^{- 4} = \frac{48 ε _{0}}{0.4} \times 1 0^{- 4} = 1.2 ε_{0} \times 1 0^{- 2}

Actually, working in cm:

C_{eff} = 1.2 ε_{0}

(in appropriate units).

Energy stored:

U = \frac{1}{2} C_{eff} V^{2} = \frac{1}{2} (1.2 ε_{0}) (20)^{2} = 240 ε_{0} J

Answer: 240

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Integer answer

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JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Capacitance

Solution

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