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28 August, 14:27

Two particles each have the same mass but particle #1 has four times the charge of particle #2. Particle #1 is accelerated from rest through a potential difference of 10 V and attains speed v. Particle #2 is accelerated from rest also through a potential difference of 10 V. What speed does particle #2 attain?

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  1. 28 August, 15:45
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    v_2 = 2*v

    Explanation:

    Given:

    - Mass of both charges = m

    - Charge 1 = Q_1

    - Speed of particle 1 = v

    - Charge 2 = 4*Q_1

    - Potential difference p. d = 10 V

    Find:

    What speed does particle #2 attain?

    Solution:

    - The force on a charged particle in an electric field is given by:

    F = Q*V / r

    Where, r is the distance from one end to another.

    - The Net force acting on a charge accelerates it according to the Newton's second equation of motion:

    F_net = m*a

    - Equate the two expressions:

    a = Q*V / m*r

    - The speed of the particle in an electric field is given by third kinetic equation of motion.

    v_f^2 - v_i^2 = 2*a*r

    Where, v_f is the final velocity,

    v_i is the initial velocity = 0

    v_f^2 - 0 = 2*a*r

    Substitute the expression for acceleration in equation of motion:

    v_f^2 = 2 * (Q*V / m*r) * r

    v_f^2 = 2*Q*V / m

    v_f = sqrt (2*Q*V / m)

    - The velocity of first particle is v:

    v = sqrt (20*Q / m)

    - The velocity of second particle Q = 4Q

    v_2 = sqrt (20*4*Q / m)

    v_2 = 2*sqrt (20*Q / m)

    v_2 = 2*v
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