First slide

Capacitance

Question

Thee plates A, B, C each of area 50 cm² have separation 3 mm between A and B and 3 mm between B and C fig. The energy stored when the plates are fully charged is

Moderate

A

$2 μJ$
B

$1.6 μJ$
C

$5 μJ$
D

$7 μJ$

Solution

The given combination is equivalent to two capacitors connected in parallel. So, the total capacitance $= 2 ε_{0} A / d$
$∴$ Energy stored $= \frac{1}{2} (\frac{2 ε_{0} A}{d}) V^{2} = \frac{ε_{0} {AV}^{'}}{d}$
Substituting the given values, we get
Energy stored $= 1.6 μJ$

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