What must the charge (sign and magnitude) of a particle of mass 1.49 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 670 n/c ?

Answer;

= -2.18 × 10^-5 C

Explanation;

m = 1.49 × 10^-3 kg  

Take downward direction as positive.  

Fg = m g  

E = 670 N/C  

Fe = q E  

Fe + Fg = 0  

q E + m g = 0  

q = -m g/E

   = -1.49 × 10^-3 × 9.81/670

    = -2.18 × 10^-5 C  

=  -2.18 × 10^-5 C

the closer the planet is to the sun, the larger the gravity