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

\varepsilon _0a^2 \over d( 1 - \alpha a \over 2d )

\varepsilon _0a^2 \over d( 1 + \alpha a \over d )

\varepsilon _0a^2 \over d( 1 - \alpha a \over 4d )

\varepsilon _0a^2 \over d( 1 - 3\alpha a \over 2d )

Capacitor PYQ — JEE Main 2020 · 8 Jan · Shift 2 · Q37 with Solution

View written solutionFree

Correct answer: C

Model the capacitor as many parallel strips

Since the plates make a very small angle $\alpha$ , the separation between the plates varies linearly along one side.

Let the minimum separation at one end be $d$ . At a distance $x$ from that end, the separation is approximately

D(x)=d+\alpha x

because for small angle, $\tan\alpha \approx \alpha$ .

The plates are square of side $a$ , so take a thin strip of width $dx$ and breadth $a$ . Its area is

dA=a\,dx

Capacitance of the thin strip

Each strip behaves like a parallel plate capacitor:

dC=\frac{\varepsilon_0\,dA}{D(x)} =\frac{\varepsilon_0 a\,dx}{d+\alpha x}

These strips are connected in parallel, so total capacitance is

C=\int_0^a dC =\varepsilon_0 a\int_0^a \frac{dx}{d+\alpha x}

Integrate

C=\varepsilon_0 a\left[\frac{1}{\alpha}\ln(d+\alpha x)\right]_0^a =\frac{\varepsilon_0 a}{\alpha}\ln\left(\frac{d+\alpha a}{d}\right)

So,

C=\frac{\varepsilon_0 a}{\alpha}\ln\left(1+\frac{\alpha a}{d}\right)

Use small-angle / small-variation approximation

Given the angle is very small, we also take

\frac{\alpha a}{d}\ll 1

Using

\ln(1+u)\approx u-\frac{u^2}{2}

with

u=\frac{\alpha a}{d}

we get

C\approx \frac{\varepsilon_0 a}{\alpha}\left(\frac{\alpha a}{d}-\frac{1}{2}\frac{\alpha^2 a^2}{d^2}\right)

C\approx \frac{\varepsilon_0 a^2}{d}-\frac{\varepsilon_0 \alpha a^3}{2d^2}

Factor out $\dfrac{\varepsilon_0 a^2}{d}$ :

C\approx \frac{\varepsilon_0 a^2}{d}\left(1-\frac{\alpha a}{2d}\right)

Match with options

This matches Option C:

\boxed{\frac{\varepsilon_0 a^2}{d}\left(1-\frac{\alpha a}{2d}\right)}

Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.

More from Capacitor

View written solutionFree

Correct answer: C

Model the capacitor as many parallel strips

Since the plates make a very small angle $\alpha$ , the separation between the plates varies linearly along one side.

Let the minimum separation at one end be $d$ . At a distance $x$ from that end, the separation is approximately

D(x)=d+\alpha x

because for small angle, $\tan\alpha \approx \alpha$ .

The plates are square of side $a$ , so take a thin strip of width $dx$ and breadth $a$ . Its area is

dA=a\,dx

Capacitance of the thin strip

Each strip behaves like a parallel plate capacitor:

dC=\frac{\varepsilon_0\,dA}{D(x)} =\frac{\varepsilon_0 a\,dx}{d+\alpha x}

These strips are connected in parallel, so total capacitance is

C=\int_0^a dC =\varepsilon_0 a\int_0^a \frac{dx}{d+\alpha x}

Integrate

C=\varepsilon_0 a\left[\frac{1}{\alpha}\ln(d+\alpha x)\right]_0^a =\frac{\varepsilon_0 a}{\alpha}\ln\left(\frac{d+\alpha a}{d}\right)

So,

C=\frac{\varepsilon_0 a}{\alpha}\ln\left(1+\frac{\alpha a}{d}\right)

Use small-angle / small-variation approximation

Given the angle is very small, we also take

\frac{\alpha a}{d}\ll 1

Using

\ln(1+u)\approx u-\frac{u^2}{2}

with

u=\frac{\alpha a}{d}

we get

C\approx \frac{\varepsilon_0 a}{\alpha}\left(\frac{\alpha a}{d}-\frac{1}{2}\frac{\alpha^2 a^2}{d^2}\right)

C\approx \frac{\varepsilon_0 a^2}{d}-\frac{\varepsilon_0 \alpha a^3}{2d^2}

Factor out $\dfrac{\varepsilon_0 a^2}{d}$ :

C\approx \frac{\varepsilon_0 a^2}{d}\left(1-\frac{\alpha a}{2d}\right)

Match with options

This matches Option C:

\boxed{\frac{\varepsilon_0 a^2}{d}\left(1-\frac{\alpha a}{2d}\right)}

Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.

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