Question

What must the charge (sign and magnitude) of a particle of mass 1.43 gg be for it to remain stationary when placed in a downward-directed electric field of magnitude 660 N/CN/C

Answer 1

The charge is 21.2μC

Step-by-step explanation:

Given that,

Mass = 1.43 g

Electric field = 660 N/C

We need to calculate the charge

Using formula of force

`F=Eq`

`mg=Eq`

`q=(mg)/(E)`

Put the value into the formula

`q=(1.43*10^(-3)*9.8)/(660)`

`q=0.00002123\ C`

`q=21.2*10^(-6)\ C`

`q=21.2\ \mu C`

Hence, The charge is 21.2μC