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a small particle carrying a negative charge of 1.6*10^-19 c is suspended in equilibrium between the horizontal metal plates 5 cms apart, having a potential difference of 3000v across them. find the mass of the particle?

Arun
25763 Points
3 years ago
Dear student
E = dV/dr = 3000/5 * 10^-2 = 6 * 104 volt /m
Now qE = mg
m = q E /g = 1.6 * 10^-19* 6 * 10^4 /9.8
= 9.8 * 10^-16 kg
Saurav
224 Points
3 years ago
IF YOU APLLY MIND YOU WON’T NEED TO REMEMBER FORMULAS:
E = dV/dr
qE = mg  (THIS IS OBVIOUSLY FOR  ABOVE GIVEN QUESTION AND NOT GENERAL)
Gitanjali Rout
184 Points
2 years ago
= dV/dr = 3000/5 * 10^-2 = 6 * 104 volt /m
Now qE = mg
m = q E /g = 1.6 * 10^-19* 6 * 10^4 /9.8
= 9.8 * 10^-16 kg
E = dV/dr
qE = mg  (THIS IS OBVIOUSLY FOR  ABOVE GIVEN QUESTION AND NOT GENERAL).
E = dV/dr = 3000/5 * 10^-2 = 6 * 104 volt /m
Now qE = mg
m = q E /g = 1.6 * 10^-19* 6 * 10^4 /9.8
= 9.8 * 10^-16 kg.