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A capacitor is made of two square plates each of side ‘a’ making a very small angle  α between them, as shown in figure. The capacitance will be close to :

a
∈0a2d1+αad
b
∈0a2d1−αa4d
c
∈0a2d1−αa2d
d
∈0a2d1−3αa2d

detailed solution

Correct option is C

⇒Select in small strip of dx, at a distance x from left.​⇒tanα=yx⇒y=xtanα=xαFor small angles tanα=α​⇒distance between plates at a distance x=d+xα⇒Capacitance of this element can be written as, ​⇒dC=ε0dAd+αx⇒dC=ε0adxd+αx      ​⇒All such elements are connected in parallel, integration will give total capacitance ⇒∫0adC=∫0aε0adxd+αx ⇒C= ε0a∫0adxd+αx​⇒C=ε0aαlnd+αx0a ∵ ∫1axdx=logaxddxax+C ⇒C=ε0aαln1+αad​⇒C≃ε0a2d1−αa2d            ln1+x =x1−x2

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