A small object of mass 4.10 g and charge −18.4 μc is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. what are the magnitude and direction of the electric field?

Respuesta :

force due to electric field is balanced by the force due to its own weight

[tex]F_g = F_e[/tex]

[tex]mg = qE[/tex]

here

m = 4.10 g

[tex]q = -18.4 \mu C[/tex]

now from the above equation we will have

[tex]4.10*10^{-3} * 9.8 = 18.4 * 10^{-6} * E[/tex]

[tex]E = 2183.7 N/C[/tex]

its direction is downwards along the direction of gravity as its a negative charge given here.