A particle with a mass of 1 g and a charge of 2.5 × 10^ -4 C is released from rest within an electric field of 1.2 × 10^4 N C^ -1 . Determine the speed of the particle after it has travelled a distance of 40 cm .

Correct answer: A. 49.0 m s^ -1 Given, mass of the particle, m = 1 g = 10^ -3 kg Charge, q = 2.5 × 10^ -4 C Electric field, E = 1.2 × 10^4 N C^ -1 Distance travelled, s = 40 cm = 0.4 m The force experienced by the particle in the electric field is: F = qE = (2.5 × 10^ -4 ) × (1.2 × 10^4) = 3.0 N The acceleration of the particle is: a = (F)/(m) = 3.0 10^ -3 = 3000 m s^ -2 Using the equation of motion v^2 = u^2 + 2as with initial velocity u = 0 : v^2 = 0 + 2 × 3000 × 0.4 = 2400 v = 2400 = 20 6 48.99 m s^ -1 49.0 m s^ -1

Concepts Of Physics MCQ Edition [Volume 2]

Exam: Concepts Of Physics MCQ Edition [Volume 2]
Subject: Physics
Chapter: Electric Field and Potential

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Electric Field and Potential

A particle with a mass of 1 g and a charge of 2.5 × 10^ -4 C is...

A particle with a mass of

1 \text{ g}

and a charge of

2.5 \times 10^{-4} \text{ C}

is released from rest within an electric field of

1.2 \times 10^4 \text{ N C}^{-1}

. Determine the speed of the particle after it has travelled a distance of

40 \text{ cm}

Options

A $49.0 \text{ m s}^{-1}$
B $34.6 \text{ m s}^{-1}$
C $24.5 \text{ m s}^{-1}$
D $98.0 \text{ m s}^{-1}$

Correct answer

A. $49.0 \text{ m s}^{-1}$

Step-by-step solution

Given, mass of the particle,

m = 1 \text{ g} = 10^{-3} \text{ kg}

Charge,

q = 2.5 \times 10^{-4} \text{ C}

Electric field,

E = 1.2 \times 10^4 \text{ N C}^{-1}

Distance travelled,

s = 40 \text{ cm} = 0.4 \text{ m}

The force experienced by the particle in the electric field is:

F = qE = (2.5 \times 10^{-4}) \times (1.2 \times 10^4) = 3.0 \text{ N}

The acceleration of the particle is:

a = \dfrac{F}{m} = \dfrac{3.0}{10^{-3}} = 3000 \text{ m s}^{-2}

Using the equation of motion

v^2 = u^2 + 2as

with initial velocity

u = 0

v^2 = 0 + 2 \times 3000 \times 0.4 = 2400

v = \sqrt{2400} = 20\sqrt{6} \approx 48.99 \text{ m s}^{-1} \approx 49.0 \text{ m s}^{-1}

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