A Levels Physics

A tiny oil droplet with mass 6.9 × 10-13 kg is at rest in an electric field of electric field strength
2.1 x 10'NC", as shown.
horizontal
plate
oil droplet
The weight of the droplet is exactly balanced by the electrical force on the droplet.
What is the charge on the droplet?
A 3.3 x 10-20C
-20
B
-3.3 x 10 2° C
с 3.2х 1019 С
D -3.2 x 10-19C


Choice D is correct as proven by a series of steps stated below.

E (Electric Field Strength) = F (Force) / Q (Charge)

E = F/Q

F = EQ

F = 2.1 x 107 Q

Electric Force on Droplet (EQ) = Weight of Droplet (mg)

2.1 x 107 Q = (6.9 x 10-13)(9.81)

Q = [ (6.9 x 10-13)(9.81) ] / 2.1 x 107

Q = 3.2 x 10-19 C (Value without the sign of charge)

Determining the sign of charge

Electric Field lines go from the positive plate to the negative plate therefore the top plate is the positive plate.

The weight of the droplet is exerted downwards therefore electric force on the droplet must be exerted upwards.

An upwards electric force will mean the droplet is of a negative charge (Q =- 3.2 x 10-19 C) in order to attract the positive plate. 

Choice A is incorrect because it states the value of charge as 3.3 x 10-20 C (Incorrect value & sign of charge).

Choice B is incorrect because it states the value of charge as -3.3 x 10-20 C (Correct sign of charge but incorrect value) 

Choice C is incorrect because it states the value of charge as 3.2 x 10-19 C (Correct value but incorrect sign of charge)  

Electric Fields | 2014 - M/J - 9702/12 | Q12