JEE Main 2025 Jan 24 Shift 1, Physics Q18: Capacitance For Geometrical Figures With Dielectric

JEE Main2025Jan 24, Shift 1Physics

A parallel plate capacitor was made with two rectangular plates, each with a length of $l = 3$ cm and breath of $b = 1$ cm. The distance between the plates is 3 µm. Out of the following, which are the ways to increase the capacitance by a factor of 10 ?

A. $l = 30$ cm, $b = 1$ cm, $d = 1$ µm

B. $l = 3$ cm, $b = 1$ cm, $d = 30$ µm

C. $l = 6$ cm, $b = 5$ cm, $d = 3$ µm

D. $l = 1$ cm, $b = 1$ cm, $d = 10$ µm

E. $l = 5$ cm, $b = 2$ cm, $d = 1$ µm

Choose the correct answer from the options given below :

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

Solution

For a parallel plate capacitor, $C = \frac{ε _{0} A}{d} = \frac{ε _{0} l b}{d}$ .

Initial capacitance: $C_{0} = \frac{ε _{0} \cdot 3 \cdot 1}{3} = ε_{0}$ (in cm/µm units; only the ratio matters).

Need a factor of 10, i.e. $\frac{l b}{d} = 30$ (in the same units).

A. $\frac{30 \cdot 1}{1} = 30$ . But $\frac{l b}{d}$ ratio gives $C / C_{0} = \frac{30/1}{3/3} = 30$ . That's a factor of 30, not 10. $\times$

(Recomputing: $C / C_{0} = \frac{( l b ) _{new} / ( d ) _{new}}{( l b ) _{old} / ( d ) _{old}} = \frac{30 \cdot 1/1}{3 \cdot 1/3} = \frac{30}{1} = 30$ .)

B. $\frac{3 \cdot 1/30}{3 \cdot 1/3} = \frac{0.1}{1} = 0.1$ . $\times$

C. $\frac{6 \cdot 5/3}{3 \cdot 1/3} = \frac{10}{1} = 10$ . $✓$

D. $\frac{1 \cdot 1/10}{1} = 0.1$ . $\times$

E. $\frac{5 \cdot 2/1}{1} = 10$ . $✓$

Cases C and E give the desired factor of 10.

$∴$ Option (1).

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JEE Main 2025 Jan 24 Shift 1, Physics Q18: Capacitance For Geometrical Figures With Dielectric

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