Question

Пылинка массой 20 мкг, несущая на себе заряд q = –40 нКл, влетела в электрическое поле в направлении силовых линий. После прохождения разности потенциалов 200 В пылинка имела скорость 10 м/с. Определить скорость пылинки до того, как она влетела в поле.

Answer 1

M = 20 мкг = 20 × 10^-9 кг
q = -40 нКл = -40 × 10^-9 Кл
U = 200 В
v = 10 м/с

v0 - ?

Решение:
A = Uq,
A = 200 × -40 × 10^-9 = -8 × 10^-6 (Дж)
A = E(к) => A = mv²/2 - mv0²/2 => mv²/2 - mv0²/2 = A => - v0² = 2A / m²v²,
- v0² = 2 × -8 × 10^-6 / (20 × 10^-9)² × 10² = -5^8 × 2^10,
- v0² = -5^8 × 2^10 => v0² = 5^8 × 2^10,
v0 = √(5^8 × 2^10) = 5^4 × 2^5 = 625 × 32 = 2 × 10^4 (м/с).

Ответ: 2 × 10^4 м/с.

Answer 2

Calculation of the Dust Particle's Initial Velocity

To determine the initial velocity of the dust particle before it entered the electric field, we can use the principle of conservation of energy. The dust particle initially has a charge of q = -40 nC and a mass of m = 20 μg. It enters the electric field with a potential difference of ΔV = 200 V and a final velocity of v = 10 m/s.

The potential difference across the electric field is given by the equation:

ΔV = q * V / m

where ΔV is the potential difference, q is the charge, V is the voltage, and m is the mass of the particle.

Rearranging the equation, we can solve for the initial velocity:

V = ΔV * m / q

Substituting the given values:

V = (200 V) * (20 μg) / (-40 nC)

Converting the units to SI:

V = (200 V) * (20 * 10^(-9) kg) / (-40 * 10^(-9) C)

Simplifying the equation:

V = -100 m/s

Therefore, the initial velocity of the dust particle before it entered the electric field was -100 m/s.