Physics, asked by financeexpert2566, 1 year ago

A charged dust particle of radius 5 107 m is located in a horizontal electric field having an intensity of 3.14 105 n/c. The surrounding medium is air with coefficient of viscosity = 1.6 105 n-sm2. If the particle moves with a uniform horizontal speed of 0.01 ms1 (neglect gravitational effects) the number of electrons on it will be

Answers

Answered by tiwaavi
6

Answer ⇒ No. of electrons are 3.06 × 10¹⁸

Explanation ⇒

Given, Radius = 5.107 m [Its given confusing, i think it will be only correct, and as per as question makes its value and follow my method to enjoy correct answer.]

E = 3.14 × 10⁵ N/C [Also given confusing, but its only correct.]

η = 1.6 × 10⁵ Nsm⁻²

v = 0.01 ms⁻¹.

Under equilibrium,

Electrostatic force = Viscous force.

∴ qE = 6πηrv

∴ q × 3.14 × 10⁵ =  6 × 3.14 × 1.6 × 10⁵ × 5.107  × 0.01

q = 0.49 C.

Now, Charge = ne

∴ n = Charge/e

∴ n = (0.49/1.6) × 10¹⁹

∴ n = 3.06 × 10¹⁸.

Hence, no of electrons are 3.06 × 10¹⁸.

Hope it helps.

Answered by Anonymous
5

Answer ⇒ No. of electrons are 3.06 × 10¹⁸

Explanation ⇒

Given, Radius = 5.107 m [Its given confusing, i think it will be only correct, and as per as question makes its value and follow my method to enjoy correct answer.]

E = 3.14 × 10⁵ N/C [Also given confusing, but its only correct.]

η = 1.6 × 10⁵ Nsm⁻²

v = 0.01 ms⁻¹.

Under equilibrium,

Electrostatic force = Viscous force.

∴ qE = 6πηrv

∴ q × 3.14 × 10⁵ =  6 × 3.14 × 1.6 × 10⁵ × 5.107  × 0.01

q = 0.49 C.

Now, Charge = ne

∴ n = Charge/e

∴ n = (0.49/1.6) × 10¹⁹

∴ n = 3.06 × 10¹⁸.

Hence, no of electrons are 3.06 × 10¹❤️

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