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

The value is
`q_2 = 6.1 *10^(-6) \ C`

Step-by-step 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

`( 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

`( 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

`( 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

`(2m_2 v^2_1 )/( (1)/(2) * m_1 * v_1^2) = (q_2 \Delta V )/( q_1 * \Delta V)`

=>
`q_2 = (4m_2 * q_1 )/(m_1)`

=>
`q_2 = (4* 0.002 * 3.8 *10^(-6) )/(0.005)`

=>
`q_2 = 6.1 *10^(-6) \ C`

=>
`q_2 = 6.1 *10^(-6) \ C`