JEE Main 2025 Jan 28 Shift 2, Physics Q2: Energy Stored In A Capacitor

JEE Main2025Jan 28, Shift 2Physics

A parallel plate capacitor of capacitance $1 μ$ F is charged to a potential difference of 20 V. The distance between plates is $1 μ$ m. The energy density between plates of capacitor is :

A
$1.8 \times 1 0^{3}$ J/m $^{3}$
B
$2 \times 1 0^{- 4}$ J/m $^{3}$
C
$2 \times 1 0^{2}$ J/m $^{3}$
D
$1.8 \times 1 0^{5}$ J/m $^{3}$

Solution

Electric field between the plates: $E = \frac{V}{d} = \frac{20}{1 \times 1 0 ^{- 6}} = 2 \times 1 0^{7}$ V/m.

Energy density of an electric field: $u = \frac{1}{2} ε_{0} E^{2}$ .

With $ε_{0} = 8.85 \times 1 0^{- 12}$ F/m:

$u = \frac{1}{2} (8.85 \times 1 0^{- 12}) (2 \times 1 0^{7})^{2} = \frac{1}{2} (8.85 \times 1 0^{- 12}) (4 \times 1 0^{14}) \approx 1.77 \times 1 0^{3}$ J/m $^{3}$ .

Closest to $1.8 \times 1 0^{3}$ J/m $^{3}$ .

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JEE Main 2025 Jan 28 Shift 2, Physics Q2: Energy Stored In A Capacitor

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