contestada

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

Respuesta :

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

[tex]F=Eq[/tex]

[tex]mg=Eq[/tex]

[tex]q=\dfrac{mg}{E}[/tex]

Put the value into the formula

[tex]q=\dfrac{1.43\times10^{-3}\times9.8}{660}[/tex]

[tex]q=0.00002123\ C[/tex]

[tex]q=21.2\times10^{-6}\ C[/tex]

[tex]q=21.2\ \mu C[/tex]

Hence, The charge is 21.2μC