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Question: A charged dust particle of radius 5 × 10⁻⁷ m is located in a horizontal electric field having an int...

A charged dust particle of radius 5 × 10⁻⁷ 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 η = 1.6 × 10⁻⁵ N-sm⁻². If the particle moves with a uniform horizontal speed of 0.1 ms⁻¹ (neglect gravitational effects), the number of electrons on it will be

A

300

B

250

C

200

D

150

Answer

300

Explanation

Solution

The charged dust particle is in a horizontal electric field and moves with a uniform horizontal speed. Since the velocity is uniform, the net horizontal force on the particle is zero. The forces acting on the particle in the horizontal direction are the electric force and the viscous drag force due to the air. We are neglecting gravitational effects, so we only consider horizontal forces.

The electric force on the charged particle with charge qq in an electric field EE is given by Fe=∣q∣EF_e = |q|E.

The viscous drag force on a spherical particle of radius rr moving with speed vv in a medium with viscosity η\eta is given by Stokes' law: Fd=6πηrvF_d = 6\pi\eta rv.

Since the particle moves with uniform speed, the electric force is balanced by the viscous drag force: Fe=FdF_e = F_d ∣q∣E=6πηrv|q|E = 6\pi\eta rv

Let nn be the number of excess electrons on the dust particle. The charge of the particle is q=−neq = -ne, where ee is the magnitude of the charge of an electron (e=1.6×10−19e = 1.6 \times 10^{-19} C). So, ∣q∣=∣−ne∣=ne|q| = |-ne| = ne.

Substituting this into the force balance equation: neE=6πηrvneE = 6\pi\eta rv

We need to find the number of electrons, nn. Solving for nn: n=6πηrveEn = \frac{6\pi\eta rv}{eE}

We are given the following values:

  • Radius of the particle, r=5×10−7r = 5 \times 10^{-7} m
  • Electric field intensity, E=3.14×105E = 3.14 \times 10^5 N/C
  • Coefficient of viscosity of air, η=1.6×10−5\eta = 1.6 \times 10^{-5} N-sm−2^{-2}
  • Uniform horizontal speed, v=0.1v = 0.1 ms−1^{-1}
  • Magnitude of the charge of an electron, e=1.6×10−19e = 1.6 \times 10^{-19} C

We can use π≈3.14\pi \approx 3.14.

Substituting the given values into the formula for nn: n=6×3.14×(1.6×10−5 N-sm−2)×(5×10−7 m)×(0.1 ms−1)(1.6×10−19 C)×(3.14×105 N/C)n = \frac{6 \times 3.14 \times (1.6 \times 10^{-5} \text{ N-sm}^{-2}) \times (5 \times 10^{-7} \text{ m}) \times (0.1 \text{ ms}^{-1})}{(1.6 \times 10^{-19} \text{ C}) \times (3.14 \times 10^5 \text{ N/C})}

n=6×3.14×1.6×5×0.1×10−5×10−71.6×3.14×10−19×105n = \frac{6 \times 3.14 \times 1.6 \times 5 \times 0.1 \times 10^{-5} \times 10^{-7}}{1.6 \times 3.14 \times 10^{-19} \times 10^5}

Cancel out the common terms 3.143.14 and 1.61.6 from the numerator and the denominator: n=6×5×0.1×10−5×10−710−19×105n = \frac{6 \times 5 \times 0.1 \times 10^{-5} \times 10^{-7}}{10^{-19} \times 10^5}

Simplify the numerical part and the powers of 10: 6×5×0.1=30×0.1=36 \times 5 \times 0.1 = 30 \times 0.1 = 3 10−5×10−7=10−1210^{-5} \times 10^{-7} = 10^{-12} 10−19×105=10−1410^{-19} \times 10^5 = 10^{-14}

So, n=3×10−1210−14n = \frac{3 \times 10^{-12}}{10^{-14}} n=3×10−12−(−14)n = 3 \times 10^{-12 - (-14)} n=3×10−12+14n = 3 \times 10^{-12 + 14} n=3×102n = 3 \times 10^2 n=300n = 300

The number of electrons on the dust particle is 300.