Question

A 5.0-g object carries a net charge of 3.8 micro-Coulomb. It acquires a speed v when accelerated from rest through a potential difference V. A 2.0-g object acquires twice the speed under the same circumstances. What is its charge?

Answer 1

Answer:

The value is  $q_2 = 6.1 *10^{-6} \ C$

Explanation:

From the question we are told that

     The mass of the object is  $m_1 = 5.0 \ g = 0.005 \ kg$

     The net charge is  $q = 3.8 \mu C= 3.8 *10^{-6} \ C$

     The mass of the second object is   $m_1 = 2.0 \ g = 0.002 \ kg$

Generally for the first object the potential energy gained at the end of its acceleration is equal to its  kinetic energy

i.e

            $\frac{ 1}{ 2} * m_1 * v^2_1 = q_1 * \Delta V \ \ --- (1)$

Here $v_1$ is the velocity of the first object

        $\Delta V$ is the potential difference through which it is accelerated  

Generally for the second object the potential energy gained at the end of its acceleration is equal to its  kinetic energy

i.e

            $\frac{ 1}{ 2} * m_2 * v^2_2 = q_2 * \Delta V$

Here $v_2$ is the velocity of the second object and from the question it is

        $v_2 = 2 v_1$

        $\Delta V$ is the potential difference through which it is accelerated        

So

          $\frac{ 1}{ 2} * m_2 * (2v)^2_1 = q_2 * \Delta V$

=>        $m_2 * 2v^2_1 = q_2 * \Delta V \ \ ---(2)$

Generally dividing equation 2 by equation 1

            $\frac{2m_2 v^2_1 }{ \frac{1}{2} * m_1 * v_1^2} = \frac{q_2 \Delta V }{ q_1 * \Delta V}$

=>         $q_2 = \frac{4m_2 * q_1 }{m_1}$

=>         $q_2 = \frac{4* 0.002 * 3.8 *10^{-6} }{0.005}$

=>         $q_2 = 6.1 *10^{-6} \ C$

=>         $q_2 = 6.1 *10^{-6} \ C$