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Capacitor question

2015 · Q111
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Capacitor question

2015 · Q111

NEETPhysicsCapacitorMCQ+4 / −1
A parallel plate air capacitor has capacity C, distance of separation between plates is d and potential difference VVV is applied between the plates. Force of attraction between the plates of the parallel plate air capacitor is
  1. A
    CV2d{{C{V^2}} \over d}dCV2​
  2. B
    C2V22d2{{{C^2}{V^2}} \over {2{d^2}}}2d2C2V2​
  3. C
    C2V22d{{{C^2}{V^2}} \over {2d}}2dC2V2​
  4. D
    CV22d{{C{V^2}} \over {2d}}2dCV2​
View written solutionFree

Correct answer: D

Force of attraction between the plates, F = qE

=q×σ2∈0=qq2A∈0 = q \times {\sigma \over {2{ \in _0}}} = q{q \over {2A{ \in _0}}}=q×2∈0​σ​=q2A∈0​q​

=q22(∈0Ad)×d=C2V22cd=CV22d = {{{q^2}} \over {2\left( {{{{ \in _0}A} \over d}} \right) \times d}} = {{{C^2}{V^2}} \over {2cd}} = {{C{V^2}} \over {2d}}=2(d∈0​A​)×dq2​=2cdC2V2​=2dCV2​

Here, C=∈0AdC = {{{ \in _0}A} \over d}C=d∈0​A​, q=CVq = CVq=CV, A = area

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