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NVAMediumJEE 2023Capacitors & Dielectrics

JEE Physics 2023 Question with Solution

As shown in the figure, two parallel plate capacitors having equal plate area of 200 cm2200 \, \text{cm}^2 are joined in such a way that α≠β\alpha \neq \beta. The equivalent capacitance of the combination is xε0x \varepsilon_0 F. The value of xx is

A capacitor arrangement showing two parallel plate capacitors with gaps labeled a and b, left spacing c = 1 mm, and total right-side separation d = 5 mm.

Answer

Correct answer:5

Step-by-step solution

Standard Method

Given: Two parallel plate capacitors have equal plate area A=200 cm2A = 200 \, \text{cm}^2. From the figure, the plate separations are aa and bb with a≠ba \neq b, total separation d=5 mmd = 5 \, \text{mm}, and one spacing is c=1 mmc = 1 \, \text{mm}.

Find: The value of xx if the equivalent capacitance is Ceq=xε0C_{eq} = x \varepsilon_0 F.

Using the diagram,

b=5−a−1=(4−a) in mmb = 5 - a - 1 = (4-a) \text{ in mm}

For each capacitor,

C=ε0AdC = \frac{\varepsilon_0 A}{d}

Since the capacitors are in series,

1Ceq=1C1+1C2\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}

Substituting the two separations aa and (4−a)(4-a),

1Ceq=aε0A+4−aε0A=4 mmε0A\frac{1}{C_{eq}} = \frac{a}{\varepsilon_0 A} + \frac{4-a}{\varepsilon_0 A} = \frac{4 \, \text{mm}}{\varepsilon_0 A}

Therefore,

Ceq=ε0A4 mmC_{eq} = \frac{\varepsilon_0 A}{4 \, \text{mm}}

Now,

A=200 cm2=200×10−4 m2A = 200 \, \text{cm}^2 = 200 \times 10^{-4} \, \text{m}^2

So the equivalent capacitance evaluates to

Ceq=5ε0 faradC_{eq} = 5\varepsilon_0 \text{ farad}

Hence, x=5x = 5.

Therefore, the required value is 55.

Common mistakes

  • Treating the two capacitors as parallel instead of series. That is incorrect because the solution explicitly uses the series relation 1Ceq=1C1+1C2\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}. First identify the actual connection from the diagram before applying any formula.

  • Using the total distance 5 mm5 \, \text{mm} directly as one plate separation. This is wrong because the diagram gives b=5−a−1=4−ab = 5-a-1 = 4-a mm after subtracting the other marked gap. Always account for all labeled distances correctly.

  • Forgetting to convert area from cm2\text{cm}^2 to m2\text{m}^2. The capacitance formula must be used in SI units. Convert 200 cm2200 \, \text{cm}^2 to 200×10−4 m2200 \times 10^{-4} \, \text{m}^2 before substitution.

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