How many electrons should be removed from a coin of mass 1.6g so that it may just flow in an electric field of intensity 10 ki power 9 N/C , directed upward.
Venkateshjha:
answer is 10^17 electrons
mg = eq
q = ne
n = mg/e
putting the values we get n = 10^17
oh.... E = 10^9
then n = 10^8
9.8
that also works
9.8×10^7
Answers
Answered by
97
Given,
Mass of coin = 1.6 g =
As we know,
E = ---------------- (1)
Where,
E = intensity of electric field
F = force
Q = charge
Q = ne ,
On putting the value ,
We get,
F = ne E
F = mg
On equating both the values of F we get,
M g = n e E
= n
On solving we get ,
Number of electrons (n) =
Thank you!
My pleasure :)
wah madam ji
thanka :)
hiiii
hiiiiii
wat you have dis many dots
matlab
wonderful ans siso❣✔✔
thank u sista ♥
Answered by
3
Answer:
Mass of coin = 1.6 g = 1.6*10^{-3}kg1.6∗10
−3
kg
As we know,
E = \frac{F}{q}
q
F
---------------- (1)
Where,
E = intensity of electric field
F = force
Q = charge
Q = ne ,
On putting the value ,
We get,
F = ne E
F = mg
On equating both the values of F we get,
M g = n e E
1.6*10^{-3}*9.81.6∗10
−3
∗9.8 = n *1.6*10^{-19}*10^{9}∗1.6∗10
−19
∗10
9
On solving we get ,
Number of electrons (n) = 9.8*10^{7}9.8∗10
7
Similar questions