Physics, asked by roshanrs5711, 9 months ago

A dust particle of mass 4*10^-12 mg is suspended in air under the influence of an electric field of magnitude 50 n/c directed vertically upwards. How many electrons were removed from the neutral dust particle?

Answers

Answered by AnmolRaii
0

Dust is made of fine particles of solid matter. On Earth, it generally consists of particles in the atmosphere that come from various sources such as soil, dust lifted by wind (an aeolian process), volcanic eruptions, and pollution.

Answered by rithvik301
5

Answer:

Explanation:

The given electric field should exert an upward force on the particle so that it cancels the gravitational force which acts on the particle in the downward direction.

Since the electric field is directed upwards, therefore the particle should be positively charged so that an upward force acts on it due to the electric field.

Now, net force on the particle = Fnet = 0

Taking the upward direction as positive, we have:

                  Fnet = Fe – Fg = 0

              => qE – mg = 0

              => q = mg/E

  Substituting m=2\times 10^{-3}kg  

                     E = 4N/C and g = 10m/s2 , we get:

      q=5\times 10^{-3}C  

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