asked by crystinia, 11 days ago

# 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: mg = eq
Venkateshjha: q = ne
Venkateshjha: n = mg/e
Venkateshjha: putting the values we get n = 10^17
Venkateshjha: oh.... E = 10^9
Venkateshjha: then n = 10^8
shree5715: 9.8
Venkateshjha: that also works
Venkateshjha: 9.8×10^7

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) = crystinia: Thank you!
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mouneesh: wat you have dis many dots
sandeep8459: matlab
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3

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

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