Question

A parallel-plate capacitor is constructed of two horizontal 12.0-cm-diameter circular plates. A 1.1 g plastic bead with a charge of -5.6 nC is suspended between the two plates by the force of the electric field between them. What is the charge on the positive plate?

Answer 1

The charge on positive plate is
`-1.92 * 10^(-7)` C

Step-by-step explanation:

Given :

Diameter
`d = 0.12` m

Radius
`r = 0.06` m

Mass of bead
`= 1.1 * 10^(-3)` Kg

Charge of bead
`q = -5.6 * 10^(-9)` C

Electric field in capacitor is given by,

`E = (Q)/(\epsilon_(o) A)`

Where
`\epsilon _(o) = 8.85 * 10^(-12)`

For finding electric field in terms of force,

`E = (F)/(q)`

But
`F = mg`

`F = 1.1 * 10^(-3 ) * 9.8` (
`g = 9.8 (m)/(s^(2) )` )

`F = 10.78 * 10^(-3)` N

So electric field,
`E = (10.78 * 10^(-3) )/(-5.6 * 10^(-9) )`

`E =- 1.92 * 10^(6)`
`(N)/(C)`

Now charge on positive plate is,

`Q = E \epsilon _(o) A`

`Q = -1.92 * 10^(6) * 8.85 * 10^(-12) * \pi (0.06)^(2)`

`Q =- 1.92 * 10^(-7)` C

Therefore, the charge on positive plate is
`-1.92 * 10^(-7)` C