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Electrostatics question

2004 · Shift 0 · Q130
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Electrostatics question

2004 · Shift 0 · Q130

JEE MainPhysicsElectrostaticsMCQ+4 / −1
A charged oil drop is suspended in a uniform field of 3×104v/m3 \times {10^4}v/m3×104v/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}}\,\,kg=9.9×10−15kg and g=10 m/s2g = 10\,m/{s^2}g=10m/s2)
  1. A
    1.6×10−18 C1.6 \times {10^{ - 18}}\,C1.6×10−18C
  2. B
    3.2×10−18 C3.2 \times {10^{ - 18}}\,C3.2×10−18C
  3. C
    3.3×10−18 C3.3 \times {10^{ - 18}}\,C3.3×10−18C
  4. D
    4.8×10−18 C4.8 \times {10^{ - 18}}\,C4.8×10−18C
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