What must the charge (sign and magnitude) of a particle of mass 1.44 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 660 N/C ? Use 9.81 m/s^2 for the magnitude of the acceleration due to gravity.

Respuesta :

Answer:

Charge, [tex]q=-2.14\times 10^{-5}\ C[/tex]

Explanation:

It is given that,

Mass of the charged particle, m = 1.44 g = 0.00144 kg

Electric field, E = 660 N/C

We need to find the charge of that particle to remain stationary when placed in a downward-directed in the given electric field such that its weight is balanced by the electrostatic force i.e.

[tex]mg=qE[/tex]

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

[tex]q=\dfrac{0.00144\ kg\times 9.81\ m/s^2}{660\ N/C}[/tex]

q = 0.0000214 C

[tex]q=2.14\times 10^{-5}\ C[/tex]

Since, the electric field is acting in downward direction, so the electric force will act in opposite direction such that they are in balanced position. Hence, the charge must be negative.

i.e. [tex]q=-2.14\times 10^{-5}\ C[/tex]