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NEET 2008
Exam
NEET
Year
2008
Subject
Physics
Chapter
Capacitance

NEET 2008 · Physics · Capacitance

The energy required to charge a parallel plate condenser of plate...

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

Options

  1. A12ε0E2/Ad\frac{1}{2} \varepsilon_0 E^2 / A d
  2. Bε0E2/Ad\varepsilon_0 E^2 / A d
  3. Cε0E2Ad\varepsilon_0 E^2 A d
  4. D12ε0E2Ad\frac{1}{2} \varepsilon_0 E^2 A d

Correct answer

C. ε0E2Ad\varepsilon_0 E^2 A d

Step-by-step solution

Energy given by the cell E=CV2E=C V^2 Here, C=capacitanceofC=capacitance ofcondenser =Aε0d\frac{A \varepsilon_0}{d} V=potentialdifferenceacrosstheplatespotential difference across the plates=E dTherefore,Therefore,E=(Aε0d)(Ed)2=Aε0E2d\begin{aligned} E & =\left(\frac{A \varepsilon_0}{d}\right)(E d)^2 \\ & =A \varepsilon_0 E^2 d \end{aligned}$
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