AP EAMCET · PHYSICS · Capacitance
A parallel plate capacitor has a capacity \(80 \times 10^{-6} \mathrm{~F}\), when air is present between its plates. The space between the plates is filled with a dielectric slab of dielectric constant 20. The capacitor is now connected to a battery of \(30 \mathrm{~V}\) by wires. The dielectric slab is then removed. Then, the charge passing through the wire is
- A \(12 \times 10^{-3} \mathrm{C}\)
- B \(25.3 \times 10^{-3} \mathrm{C}\)
- C \(120 \times 10^{-3} \mathrm{C}\)
- D \(45.6 \times 10^{-3} \mathrm{C}\)
Answer & Solution
Correct Answer
(D) \(45.6 \times 10^{-3} \mathrm{C}\)
Step-by-step Solution
Detailed explanation
\(C_{\text {air }}=80 \mu \mathrm{F}\) \[ C_{\text {dielectric }}=\varepsilon_r C_{\text {air }}=1600 \mu \mathrm{F} \] Charge stored in the presence of air, \[ q_{\text {air }}=C_{\text {air }} \times V=80 \times 30 \times 10^{-6}=2400 \mu \mathrm{C} \] Charge stored in…
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