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The energy required to charge a parallel plate condenser of plate separation $d$ and plate area of cross-section $A$ such that the uniform electric field between the plates is $E$, is 

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Important Questions on Capacitance

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A 40 μF capacitor in a defibrillator is charged to 3000 V. The energy stored in the capacitor is sent through the patient during a pulse of duration 2 ms. The power delivered to the patient is

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If there are n capacitors of capacitance C in parallel connected to V volt source, then the energy stored is equal to:
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A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and the work done by the battery will be 
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A parallel plate capacitor with air between the plates has a capacitance of 9 pF. The separation between its plates is d. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant k1=3 and thickness d/3 while the other one has dielectric constant k2=6 and thickness 2d/3. Capacitance of the capacitor is now
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If the energy of a capacitor of capacitance C=2 μF is 0.16 J, then its potential difference will be
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A capacitor of 6 μF is charged to such an extent that the potential difference between the plates becomes 50 V. The work done in this process will be,
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Two identical capacitors have the same capacitance C. One of them is charged to potential V1 and the other to V2. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is,
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A capacitor is connected to a cell of emf E having some internal resistance r. The potential difference across(in steady state),