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
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.
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❤️