Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
MHT CET2008PhysicsCapacitance

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

Options

  1. A1 2 ₀ E² / A d
  2. B₀ E² / A d
  3. C₀ E² A d
  4. D1 2 ₀ E² A d

Correct answer

C. ₀ E² A d

Step-by-step solution

Energy given by the cell aligned & E=C V² Here, C &= capacitance of condenser &= A ₀ d aligned V= potential difference across the plates =E d Therefore, E= ( A ₀ d )(E d)²=A ₀ E² d

Practice Capacitance on Quantrex Academy →

More from Capacitance

A capacitor of capacitance 8 F is fully charged by connecting it to a source of 200V. It is then disconnected from the supply and connected to an uncharged capacitor of capacitance 2026What is the charge on 15 F in the circuit given? 2026A parallel plate capacitor has a uniform electric field ' E ' in the space between the plates. If the distance between the plates is ' d ' and area of each plate is ' A ', the ener 2026In the circuit shown in the figure, the potential difference across the 4 F capacitor is 2026A parallel plate air capacitor having area of each plate ' A ' and the distance between the plates ' d ' has uniform electric field E in the space between the plates. The energy st 2026In a parallel plate air capacitor of plate separation ' d ', a dielectric slab of thickness ' t ' is introduced between the plates. The capacitance becomes one-third of the origina 2026A parallel plate air capacitor has capacity 'C' farad, potential 'V' volt and energy 'E' joule. When the gap between the plates is completely filled with dielectric 2026In an oscillating LC circuit the maximum charge on the capacitor is Q. When the energy is stored equally between the electric and magnetic fields, the charge on the capacitor (q) i 2026 Full Capacitance list All MHT CET PYQs