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Capacitor question

2020 · 7 Jan · Shift 1 · Q47
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Capacitor question

2020 · 7 Jan · Shift 1 · Q47

JEE MainPhysicsCapacitorMCQ+4 / −1
JEE Main 2020 (Online) 7th January Morning Slot Physics - Capacitor Question 112 English A parallel plate capacitor has plates of area A separated by distance 'd' between them. It is filled with a dielectric which has a dielectric constant that varies as k(x) = K(1 + α\alphaα x) where 'x' is the distance measured from one of the plates. If (ad) << 1, the total capacitance of the system is best given by the expression :
  1. A
    A∈0Kd(1+(αd2)2){{A{ \in _0}K} \over d}\left( {1 + {{\left( {{{\alpha d} \over 2}} \right)}^2}} \right)dA∈0​K​(1+(2αd​)2)
  2. B
    A∈0Kd(1+αd2){{A{ \in _0}K} \over d}\left( {1 + {{\alpha d} \over 2}} \right)dA∈0​K​(1+2αd​)
  3. C
    A∈0Kd(1+α2d22){{A{ \in _0}K} \over d}\left( {1 + {{{\alpha ^2}{d^2}} \over 2}} \right)dA∈0​K​(1+2α2d2​)
  4. D
    A∈0Kd(1+αd){{A{ \in _0}K} \over d}\left( {1 + \alpha d} \right)dA∈0​K​(1+αd)
View written solutionFree

Correct answer: B

  1. Model the capacitor as a series combination of thin dielectric slabs

Since the dielectric constant varies with position, k(x)=K(1+αx),0≤x≤dk(x)=K(1+\alpha x), \qquad 0\le x\le dk(x)=K(1+αx),0≤x≤d we divide the space between the plates into thin slabs of thickness dxdxdx.

For a slab at position xxx, the permittivity is ε(x)=ε0k(x)=ε0K(1+αx).\varepsilon(x)=\varepsilon_0 k(x)=\varepsilon_0 K(1+\alpha x).ε(x)=ε0​k(x)=ε0​K(1+αx).

The capacitance of a thin slab of thickness dxdxdx and area AAA is dC=ε(x)Adx.dC=\frac{\varepsilon(x)A}{dx}.dC=dxε(x)A​.

But these slabs are stacked along the field direction, so they are in series. Hence we add their inverses: d\left(\frac{1}{C}\right)=\frac{dx}{\varepsilon(x)A}= rac{dx}{\varepsilon_0 K A(1+\alpha x)}.

So, 1C=∫0ddxε0KA(1+αx).\frac{1}{C}=\int_0^d \frac{dx}{\varepsilon_0 K A(1+\alpha x)}.C1​=∫0d​ε0​KA(1+αx)dx​.

  1. Evaluate the integral

1C=1ε0KA∫0ddx1+αx.\frac{1}{C}=\frac{1}{\varepsilon_0 K A}\int_0^d \frac{dx}{1+\alpha x}.C1​=ε0​KA1​∫0d​1+αxdx​.

Using ∫dx1+αx=1αln⁡(1+αx),\int \frac{dx}{1+\alpha x}=\frac{1}{\alpha}\ln(1+\alpha x),∫1+αxdx​=α1​ln(1+αx), we get 1C=1ε0KA⋅1αln⁡(1+αd).\frac{1}{C}=\frac{1}{\varepsilon_0 K A}\cdot \frac{1}{\alpha}\ln(1+\alpha d).C1​=ε0​KA1​⋅α1​ln(1+αd).

Therefore, C=ε0KA αln⁡(1+αd).C=\frac{\varepsilon_0 K A\,\alpha}{\ln(1+\alpha d)}.C=ln(1+αd)ε0​KAα​.

  1. Use the condition αd≪1\alpha d\ll 1αd≪1

For small αd\alpha dαd, ln⁡(1+αd)≈αd−(αd)22.\ln(1+\alpha d)\approx \alpha d-\frac{(\alpha d)^2}{2}.ln(1+αd)≈αd−2(αd)2​. Thus,

Factor out αd\alpha dαd from the denominator:

=\frac{\varepsilon_0 K A}{d}\cdot \frac{1}{1-\frac{\alpha d}{2}}.$$ Now use $$\frac{1}{1-x}\approx 1+x \quad (x\ll 1),$$ so $$C\approx \frac{\varepsilon_0 K A}{d}\left(1+\frac{\alpha d}{2}\right).$$ 4. **Match with the options** This matches: $$\boxed{\frac{A\varepsilon_0 K}{d}\left(1+\frac{\alpha d}{2}\right)}$$ which is **Option B**. 5. **Option check** - **A:** correction is of order $(\alpha d)^2$ only, not the leading first-order term. - **B:** correct first-order approximation. - **C:** wrong coefficient and misses first-order term. - **D:** overestimates the first-order correction; should be $\alpha d/2$, not $\alpha d$. Hence the correct answer is **B**.
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