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JEE Main 20208 JanuaryEvening Shift
Exam
JEE Main
Year
2020
Date
8 January
Shift
Evening Shift
Subject
Physics
Chapter
Capacitance

JEE Main 2020 · 8 January · Evening Shift · Physics · Capacitance

A capacitor is made of two square plates each of side ‘ a...

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

Options

  1. A∈0a2d(1−\alphaa2d)\dfrac{{\in}_{0}{a}^{2}}{d}\left(1 - \dfrac{\alphaa}{2d}\right)
  2. B∈0a2d(1−\alphaa4d)\dfrac{{\in}_{0}{a}^{2}}{d}\left(1 - \dfrac{\alphaa}{4d}\right)
  3. C∈0a2d(1+\alphaad)\dfrac{{\in}_{0}{a}^{2}}{d}\left(1 + \dfrac{\alphaa}{d}\right)
  4. D∈0a2d(1−3\alphaa2d)\dfrac{{\in}_{0}{a}^{2}}{d}\left(1 - \dfrac{3\alphaa}{2d}\right)

Correct answer

A. ∈0a2d(1−\alphaa2d)\dfrac{{\in}_{0}{a}^{2}}{d}\left(1 - \dfrac{\alphaa}{2d}\right)

Step-by-step solution

A capacitor is made of two square plates each of side ‘ a ’ making a very small angle α between them, as
dc=∈0adxd+\alphaxdc = \dfrac{{\in}_{0}adx}{d + \alphax}
⇒c=∈0aα[In(d+\alphax)]0a⇒ c = \dfrac{{\in}_{0}a}{\alpha}{\left[In\left(d + \alphax\right)\right]}_{0}^{a}
=∈0aαIn(1+\alphaad)≈∈0a2d(1−\alphaa2b)= \dfrac{{\in}_{0}a}{\alpha}In\left(1 + \dfrac{\alphaa}{d}\right) \approx \dfrac{{\in}_{0}{a}^{2}}{d}\left(1 - \dfrac{\alphaa}{2b}\right)
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