# A charged plate holds 3 nC of charge equally distrubuted on the plates surface area. The plate is 1 cm x 1 cm. a) What is the constant electric field near the surface of this plate? b) If the plate...

A charged plate holds 3 nC of charge equally distrubuted on the plates surface area. The plate is 1 cm x 1 cm.

a) What is the constant electric field near the surface of this plate?

b) If the plate in part a) were laying flat on the table and you observe tiny charged droplets of oil levitating above the plate, what is the mass of the oil droplet if the charge on each droplet is 0.1 nC?

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### 1 Answer

a) The surface charge density of the plate is the total charge divided by the surface area: `sigma = q/A` . The area of the plate is given to be 1 square cm, or `10^-4 m^2` .

Th surface charge density is then `sigma = 3*10^-9 C/(10^-4 m^2) = 3*10^-5 C/m^2`

The electric field near the surface of the plate with the given charge density is

`E = sigma/(2epsilon_0)`

`epsilon_0` is the constant equal to `8.85 * 10^-12 N^-1 m^-2 C^2`

Plugging the values for `sigma` and `epsilon_0` into the equation above, we get the value of the electric field:

`E = (3*10^-5)/(8.85*10^-12) N/C=3.4*10^-8 N/C`

b) The electric force acting on a charged droplet of oil can be calulated as

`F = qE` , where q is the charge of the droplet and E is the electric field found above.

`F = 0.1*10^-9C * 3.4*10^-8 N/C = 3.4*10^-18 N` .

This force must balance out the force of gravity in order for the oil droplets to levitate. (We can assume the oil and the plate have like charges and the electric force on the droplet is directed upward; otherwise it would NOT be able to levitate.) Thus, the force of gravity, mg, has to equal F:

`m* 9.8 m/s^2 = 3.4*10^-18 N`

`m=(3.4*10^-18 N)/(9.8 m/s^2) =0.35 *10^-18 kg`

The mass of the droplet is `0.35*10^-18 kg` .

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