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JEE Main 2004 (AIEEE) · Motion of Charges in an Electric Field

Motion of Charges in an Electric Field question from JEE Main 2004 (AIEEE)

Q44JEE Main 2004 · AIEEEEasy

Motion of Charges in an Electric Field

A charged oil drop is suspended in a uniform field of 3×104 V/m3 \times 10^{4}\text{ V/m} so that it neither falls nor rises. The charge on the drop will be (Take the mass of the charge =9.9×10−15 kg= 9.9 \times 10^{-15}\text{ kg} and g=10 m/s2g = 10\text{ m/s}^2)

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Answer: (c)

  1. Balance: qE=mgqE = mg, so q=mgEq = \frac{mg}{E}.
  2. q=9.9×10−15×103×104=3.3×10−18 Cq = \frac{9.9 \times 10^{-15} \times 10}{3 \times 10^{4}} = 3.3 \times 10^{-18}\text{ C}
  3. So the answer is (c) 3.3×10−18 C3.3 \times 10^{-18}\text{ C}.
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