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JEE Main 2019 (8 Apr, Shift 2) · Parallel Plate Capacitor

Parallel Plate Capacitor question from JEE Main 2019 (8 Apr, Shift 2)

Q17JEE Main 2019 · 8 APR · SHIFT 2Easy

Parallel Plate Capacitor

A parallel plate capacitor has 1μF capacitance. One of its two plates is given +2μC charge and the other plate, +4μC charge. The potential difference developed across the capacitor is:-

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Answer: (a)

  1. With plate charges q1q_1, q2q_2, the inner faces carry ±q1−q22\pm\frac{q_1 - q_2}{2}.
  2. V=q1−q22C=(4−2) μC2×1 μFV = \frac{q_1 - q_2}{2C} = \frac{(4 - 2)\ \mu\text{C}}{2\times1\ \mu\text{F}}
  3. So the answer is (a) 1 V1\text{ V}.
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