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JEE Main
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Physics
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Capacitance

JEE Main · Physics · Capacitance

A parallel plate capacitor has a capacity 80 × 10^ -6 ~F , when air...

A parallel plate capacitor has a capacity 80×10−6 F80 \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 V30 \mathrm{~V} by wires. The dielectric slab is then removed. Then, the charge passing through the wire is

Options

  1. A12×10−3C12 \times 10^{-3} \mathrm{C}
  2. B25.3×10−3C25.3 \times 10^{-3} \mathrm{C}
  3. C120×10−3C120 \times 10^{-3} \mathrm{C}
  4. D45.6×10−3C45.6 \times 10^{-3} \mathrm{C}

Correct answer

D. 45.6×10−3C45.6 \times 10^{-3} \mathrm{C}

Step-by-step solution

Cair =80μFC_{\text {air }}=80 \mu \mathrm{F} C_{\text {dielectric }}=\varepsilon_r C_{\text {air }}=1600 \mu F\mathrm{F}Charge stored in the presence of air, q_{\text {air }}=C_{\text {air }} ×\timesV=80 ×\times30 ×\times10^{-6}=2400 \mu C\mathrm{C}Charge stored in presence of dielectric medium, qd=Cdielectric \timesV=1600×30×10−6=48000μC\begin{aligned}q_d=C_{\text {dielectric }} \timesV & =1600 \times30 \times10^{-6} \\ & =48000 \mu \mathrm{C} \end{aligned}When dielectric is removed, effective charge remained is q=qd−qair =(48−2.4)×10−3Cq=45.6×10−3C\begin{aligned}q & =q_d-q_{\text {air }} \\ & =(48-2.4) \times10^{-3} \mathrm{C}\\ q & =45.6 \times10^{-3} \mathrm{C} \end{aligned}
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