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JEE Main 2025 (24 Jan, Shift 1) · Parallel Plate Capacitor

Parallel Plate Capacitor question from JEE Main 2025 (24 Jan, Shift 1)

Q3JEE Main 2025 · 24 JAN · SHIFT 1Easy

Parallel Plate Capacitor

A parallel plate capacitor was made with two rectangular plates, each with a length of l=3 cml = 3\text{ cm} and breadth of b=1 cmb = 1\text{ cm}. The distance between the plates is 3 μm3\ \mu\text{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=1 μml = 30\text{ cm},\ b = 1\text{ cm},\ d = 1\ \mu\text{m}

B. l=3 cm, b=1 cm, d=30 μml = 3\text{ cm},\ b = 1\text{ cm},\ d = 30\ \mu\text{m}

C. l=6 cm, b=5 cm, d=3 μml = 6\text{ cm},\ b = 5\text{ cm},\ d = 3\ \mu\text{m}

D. l=1 cm, b=1 cm, d=10 μml = 1\text{ cm},\ b = 1\text{ cm},\ d = 10\ \mu\text{m}

E. l=5 cm, b=2 cm, d=1 μml = 5\text{ cm},\ b = 2\text{ cm},\ d = 1\ \mu\text{m}

Choose the correct answer from the options given below:

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Answer: (d)

  1. C=ε0lbd∝lbdC = \frac{\varepsilon_0 lb}{d} \propto \frac{lb}{d}. Original: 3×13=1\frac{3\times1}{3} = 1 (cm² per μm); need 10.
  2. A: 301=30\frac{30}{1} = 30; B: 330=0.1\frac{3}{30} = 0.1; C: 303=10\frac{30}{3} = 10 ✓
  3. D: 110=0.1\frac{1}{10} = 0.1; E: 101=10\frac{10}{1} = 10 ✓
  4. So the answer is (d) C and E only.
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