JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
A parallel-plate capacitor of capacitance \(40 \mu \mathrm{~F}\) is connected to a 100 V power supply. Now the intermediate space between the plates is filled with a dielectric material of dielectric constant \(\mathrm{K}=2\). Due to the introduction of dielectric material, the extra charge and the change in the electrostatic energy in the capacitor, respectively, are
- A 4 mC and 0.2 J
- B 8 mC and 2.0 J
- C 2 mC and 0.4 J
- D 2 mC and 0.2 J
Answer & Solution
Correct Answer
(A) 4 mC and 0.2 J
Step-by-step Solution
Detailed explanation
\begin{aligned} & \Delta \mathrm{q}=(\mathrm{KC}-\mathrm{C}) \mathrm{V} \\ & =40 \times 10^{-6} \times 100 \\ & =4000 \times 10^{-3}=4 \mathrm{mC} \\ & \Delta \mathrm{U}=\frac{1}{2} \mathrm{C}^{\prime} \mathrm{V}^2-\frac{1}{2} \mathrm{CV}^2=\frac{1}{2}(\mathrm{~K}-1)…
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