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?

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