Question

A small charged bead has a mass of 3.6 g. It is held in a uniform electric field E = (200,000 N/C, up). When the bead is released, it accelerates upward with an acceleration of 24 m/s^2. What is the charge on the bead?

Answer 1

The charge on the bead is
`q = 6.084 *10^(-7)\ C`

Step-by-step explanation:

From the question we are told that

The mass of the bead is
`m = 3.6 \ g = 0.0036 \ kg`

The magnitude of the electric field is
`E = 200,000 \ N/C`

The acceleration of the bead is
`a = 24 m/s^2`

Generally, the electric force on the bead is mathematically represented as

`F_ e = q E`

Where q is the charge on the bead

Now the gravitational force opposing the upward movement of the bead is mathematically represented as

`F_g = mg`

Generally the net force on the bead is mathematically represented as

`F = F_e - F_g = m* a`

=>
`qE - mg = ma`

Now substituting values

`q * 200000 - 0.0036 *9.8 = 0.0036 * 24`

`q = 6.084 *10^(-7)\ C`