SATHEE: Electrostatics Question 671

Electrostatics Question 671

Question: A charged oil drop is suspended in a uniform field of $3 \times 10^{4} V / m$ so that it neither falls nor rises. The charge on the drop will be (take the mass of the charge as $9.9 \times 10^{- 15} k g a n d g a s 10 m / s^{2}$

Options:

A) $3.3 \times 10^{- 18} C$

B) $3.2 \times 10^{- 18} C$

C) $1.6 \times 10^{- 18} C$

D) $4.8 \times 10^{- 18} C$

Show Answer

Answer:

Correct Answer: A

Solution:

[a] In equilibrium, $q E = m g$

$\Rightarrow q = 3.3 \times 10^{- 18} C$

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Electrostatics Question 671

Question: A charged oil drop is suspended in a uniform field of 3×104V/m so that it neither falls nor rises. The charge on the drop will be (take the mass of the charge as 9.9×10−15kgandgas10m/s2

Options:

Answer:

Solution:

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Question: A charged oil drop is suspended in a uniform field of $3 \times 10^{4} V / m$ so that it neither falls nor rises. The charge on the drop will be (take the mass of the charge as $9.9 \times 10^{- 15} k g a n d g a s 10 m / s^{2}$