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2. Electric potential and capacitance — Physics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 SciencePhysics2. Electric potential and capacitance1 Mark
MCQ
Energy per unit volume for a capacitor having area $A$ and separation $d$ kept at potential difference $V$ is given by
  • A
    $\frac{Q^{2}}{2 C}$
  • B
    $\frac{1}{2} C V^{2}$
  • C
    $\frac{1}{2 \varepsilon_{0}} \frac{V^{2}}{d^{2}} $
  • ✓
    $\frac{1}{2} \varepsilon_{0} \frac{V^{2}}{d^{2}}$
✓

Answer

Correct option: D.
$\frac{1}{2} \varepsilon_{0} \frac{V^{2}}{d^{2}}$
d
For a parallel plate capacitor: $\quad C =\frac{ A \varepsilon_{0}}{ d }$

Volume of capacitor $= Ad$

Energy stored in a capacitor $= E =\frac{1}{2} CV ^{2}$

Energy stored per unit volume $= E ^{\prime}=\frac{\frac{1}{2} CV ^{2}}{ Ad }$

$E ^{\prime}=\frac{1}{2} \frac{\frac{A \varepsilon_{0}}{ d } V ^{2}}{ Ad }=\frac{1}{2} \varepsilon_{0} \frac{ V ^{2}}{ d ^{2}}$

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