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

96.77 amp

explanation:

rl = 250/100 = 2.5 ohms.

egl = 250 + (100 × 0.1) + 2 = 262 v.

at 700 r.p.m., eg2 = 262 × 700/1000 = 183.4 v.

if ia is the new current, then eg2 - 2 - (ia × 0.1) = 2.5 ia

which means this will give ia = 96.77 amp.