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QJEEM20Ph4c66: JEE Main 2020 Physics - Class 12 JEEM20Ph4c66 Solution: A capacitor is made of two square plates each of side 'a'…

A capacitor is made of two square plates each of side 'a' making a very small angle α\alpha between them, as shown in figure. The capacitance will be close to : JEE Main 2020 (Online) 8th January Evening Slot Physics - Capacitor Question 103 English

Multiple Choice Options

  • A ε0a2d(1+αad){{{\varepsilon _0}{a^2}} \over d}\left( {1 + {{\alpha a} \over {d}}} \right)
  • B ε0a2d(1−αa4d){{{\varepsilon _0}{a^2}} \over d}\left( {1 - {{\alpha a} \over {4d}}} \right)
  • C ε0a2d(1−αa2d){{{\varepsilon _0}{a^2}} \over d}\left( {1 - {{\alpha a} \over {2d}}} \right)
  • D ε0a2d(1−3αa2d){{{\varepsilon _0}{a^2}} \over d}\left( {1 - {{3\alpha a} \over {2d}}} \right)

Step-by-Step Verified Solution

JEE Main 2020 (Online) 8th January Evening Slot Physics - Capacitor Question 103 English Explanation
dC = ε0adxd+αx{{{\varepsilon _0}adx} \over {d + \alpha x}}

∴\therefore C = ∫0aε0adxd+αx\int\limits_0^a {{{{\varepsilon _0}adx} \over {d + \alpha x}}}

= ε0aα[ln⁡(d+α)]0a{{{\varepsilon _0}a} \over \alpha }\left[ {\ln \left( {d + \alpha } \right)} \right]_0^a

= ε0aαln⁡[d+αad]{{{\varepsilon _0}a} \over \alpha }\ln \left[ {{{d + \alpha a} \over d}} \right]

= ε0aαln⁡[1+αad]{{{\varepsilon _0}a} \over \alpha }\ln \left[ {1 + {{\alpha a} \over d}} \right]

= ε0aα(αad−α2a2d2){{{\varepsilon _0}a} \over \alpha }\left( {{{\alpha a} \over d} - {{{\alpha ^2}{a^2}} \over {{d^2}}}} \right)

= ε0a2d(1−αa2d){{{\varepsilon _0}{a^2}} \over d}\left( {1 - {{\alpha a} \over {2d}}} \right)

Note : ln(1 + x) = x−x22+x33−...x - {{{x^2}} \over 2} + {{{x^3}} \over 3} - ...

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