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QJEEM25Ph302f: JEE Main 2025 Physics - Class 12 JEEM25Ph302f Solution: A parallel plate capacitor was made with two rectangular…

A parallel plate capacitor was made with two rectangular plates, each with a length of l=3 cml=3 \mathrm{~cm} and breath of b=1 cm\mathrm{b}=1 \mathrm{~cm} . The distance between the plates is 3μ m3 \mu \mathrm{~m} . Out of the following, which are the ways to increase the capacitance by a factor of 10 ?

A. l=30 cm, b=1 cm, d=1μ ml=30 \mathrm{~cm}, \mathrm{~b}=1 \mathrm{~cm}, \mathrm{~d}=1 \mu \mathrm{~m}

B. l=3 cm, b=1 cm, d=30μ ml=3 \mathrm{~cm}, \mathrm{~b}=1 \mathrm{~cm}, \mathrm{~d}=30 \mu \mathrm{~m}

C. l=6 cm, b=5 cm, d=3μ ml=6 \mathrm{~cm}, \mathrm{~b}=5 \mathrm{~cm}, \mathrm{~d}=3 \mu \mathrm{~m}

D. l=1 cm, b=1 cm, d=10μ ml=1 \mathrm{~cm}, \mathrm{~b}=1 \mathrm{~cm}, \mathrm{~d}=10 \mu \mathrm{~m}

E. l=5 cm, b=2 cm, d=1μ ml=5 \mathrm{~cm}, \mathrm{~b}=2 \mathrm{~cm}, \mathrm{~d}=1 \mu \mathrm{~m}

Choose the correct answer from the options given below:

Multiple Choice Options

  • A C only
  • B A only
  • C B and D only
  • D C and E only

Step-by-Step Verified Solution

The capacitance of a parallel plate capacitor is given by

C=ϵ0Ad,C = \frac{\epsilon_0 A}{d},

where:

ϵ0\epsilon_0 is the permittivity of free space,

AA is the area of a plate, and

dd is the separation between the plates.

A change in the capacitor’s dimensions will change its capacitance by a factor of

CnewCinitial=AnewAinitial⋅dinitialdnew.\frac{C_{\text{new}}}{C_{\text{initial}}} = \frac{A_{\text{new}}}{A_{\text{initial}}} \cdot \frac{d_{\text{initial}}}{d_{\text{new}}}.

The initial dimensions are:

Length: l=3 cm=0.03 m,l = 3\,\text{cm} = 0.03\,\text{m},

Breadth: b=1 cm=0.01 m,b = 1\,\text{cm} = 0.01\,\text{m},

Hence, the original area is

Ainitial=0.03×0.01=3×10−4 m2,A_{\text{initial}} = 0.03 \times 0.01 = 3 \times 10^{-4}\,\text{m}^2,

Plate separation: dinitial=3 μm=3×10−6 m.d_{\text{initial}} = 3\,\mu\text{m} = 3 \times 10^{-6}\,\text{m}.

For each modification, we compare the new factor:

Case C:

New dimensions:

l=6 cm=0.06 m,b=5 cm=0.05 m.l = 6\,\text{cm} = 0.06\,\text{m}, \quad b = 5\,\text{cm} = 0.05\,\text{m}.

New area:

Anew=0.06×0.05=3×10−3 m2.A_{\text{new}} = 0.06 \times 0.05 = 3 \times 10^{-3}\,\text{m}^2.

Ratio of areas:

AnewAinitial=3×10−33×10−4=10.\frac{A_{\text{new}}}{A_{\text{initial}}} = \frac{3 \times 10^{-3}}{3 \times 10^{-4}} = 10.

The plate separation is unchanged:

dinitialdnew=3×10−63×10−6=1.\frac{d_{\text{initial}}}{d_{\text{new}}} = \frac{3 \times 10^{-6}}{3 \times 10^{-6}} = 1.

Overall factor:

10×1=10.10 \times 1 = 10.

Case E:

New dimensions:

l=5 cm=0.05 m,b=2 cm=0.02 m.l = 5\,\text{cm} = 0.05\,\text{m}, \quad b = 2\,\text{cm} = 0.02\,\text{m}.

New area:

Anew=0.05×0.02=1×10−3 m2.A_{\text{new}} = 0.05 \times 0.02 = 1 \times 10^{-3}\,\text{m}^2.

Ratio of areas:

AnewAinitial=1×10−33×10−4≈3.33.\frac{A_{\text{new}}}{A_{\text{initial}}} = \frac{1 \times 10^{-3}}{3 \times 10^{-4}} \approx 3.33.

New plate separation:

dnew=1 μm=1×10−6 m,d_{\text{new}} = 1\,\mu\text{m} = 1 \times 10^{-6}\,\text{m},

so the ratio of separations is

dinitialdnew=3×10−61×10−6=3.\frac{d_{\text{initial}}}{d_{\text{new}}} = \frac{3 \times 10^{-6}}{1 \times 10^{-6}} = 3.

Overall factor:

3.33×3≈10.3.33 \times 3 \approx 10.

Both Case C and Case E yield a capacitance that is 10 times the original value.

Thus, the correct modifications are those in Case C and Case E.

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