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

Answer:

The charge is 21.2μC

Explanation:

Given that,

Mass = 1.43 g

Electric field = 660 N/C

We need to calculate the charge

Using formula of force

$F=Eq$

$mg=Eq$

$q=\dfrac{mg}{E}$

Put the value into the formula

$q=\dfrac{1.43\times10^{-3}\times9.8}{660}$

$q=0.00002123\ C$

$q=21.2\times10^{-6}\ C$

$q=21.2\ \mu C$

Hence, The charge is 21.2μC