Question

A 0.10 g glass bead is charged by the removal of 1.0 x 10^10 electrons. what electric field strength will cause the bead to hang suspended in the air?

Answer 1

Answer & Explanation:

To solve this problem, we need to set the gravitational force acting on the bead equal to the electric force acting on it. The bead will hang suspended in the air when these two forces are equal.

The gravitational force
`\( F_g \)` is given by:

`$$ F_g = m \cdot g $$`

where
`\( m \)` is the mass of the bead and
`\( g \)` is the acceleration due to gravity.

The electric force
`\( F_e \)` is given by:

`$$ F_e = q \cdot E $$`

where
`\( q \)` is the charge of the bead and
`\( E \)` is the electric field strength.

Setting these two equal gives:

`$$ m \cdot g = q \cdot E $$`

Solving for
`\( E \)` gives:

`$$ E = (m \cdot g)/(q) $$`

Given that the mass
`\( m \)` of the bead is 0.10 g (or 0.10/1000 kg), the acceleration due to gravity
`\( g \)` is approximately 9.8 m/s², and the charge
`\( q \)` is the charge of
`1.0 x 10^10` electrons (with the charge of one electron being approximately
`\( 1.6 * 10^(-19) \) C)`, we can substitute these values into the formula to find the electric field strength. Let's calculate that.

The electric field strength that will cause the bead to hang suspended in the air is approximately
`\(6.13 * 10^5\)` N/C (Newtons per Coulomb).