# how many electrons should be added or removed from a neutral body of mass 10 mg so that it May remain stationary in air in an electric field of strength 100 N/C directed upwards (g = 10m/s^2)

Vikas TU
14149 Points
6 years ago
Dear Student,
The given electric field ought to apply an upward constrain on the molecule so it scratchs off the gravitational compel which follows up on the molecule the descending way.
Since the electric field is coordinated upwards, along these lines the molecule ought to be decidedly charged so an upward compel follows up on it because of the electric field.
Presently, net drive on the molecule = Fnet = 0
Taking the upward course as positive, we have:
Fnet = Fe – Fg = 0
=> qE – mg = 0
=> q = mg/E
Substituting
E = 4N/C and g = 10m/s2 , we get:
q = 5mC.
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)
Rahul Kumar Bastia
15 Points
4 years ago
Fnet=Fe- mg
=>Fe = mg. Again, Fe = qE = neE
=>neE = mg
=>n = mg/eE
Given E = 100 N/C , m= 10×10–3 = 10–2 kg and g=10 m/ s²
=>n = 10–2×10/1.6 ×10–19×100
=> n = 6.25 × 10^15 electrons
As the E is in upward direction hence electrons must be removed to make it stable.
Vikhyat
15 Points
4 years ago

Fnet=Fe- mg
=>Fe = mg. Again, Fe = qE = neE
=>neE = mg
=>n = mg/eE
Given E = 100 N/C , m= 10×10–3 = 10–6 kg and g=10 m/ s²
=>n = (10–6×10) ÷(1.6 ×10–19×100)
=> n = 6.25 × 10^12 electrons
As the E is in upward direction hence electrons must be removed to make it stable.
Kiran Pawar
13 Points
4 years ago
Dear students here n is asked so one simple formula is
Charge = no.× ectron
Charge/electron =no
And as know Fe = mg
Ee= mg
E×e/charge = mg