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A charged oil drop is suspended in a uniform field of 3 10^4 ~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 and g =10 ~m / s ^2 )

Options

  1. A3.3 10⁻¹⁸ C
  2. B3.2 10⁻¹⁸ C
  3. C1.6 10⁻¹⁸ C
  4. D4.8 10⁻¹⁸ C

Correct answer

A. 3.3 10⁻¹⁸ C

Step-by-step solution

Since ball is hanging in equilibrium, force by gravity is balanced by electric force. aligned & qE = mg & q = m g E & 9.9 10⁻¹⁵ 10 3 10^4 & q =3.3 10⁻¹⁸ C aligned

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