Question

How many electrons can be removed from a piece of metal so that it has a positive charge of 8 × 10 ^ -7.​

Answer 1

Answer:

we required to remove 5×10^12 electrons which will have charge of -8×10^-7(-ve because electrons has negative charge i.e.. -1.6×10^-19 C).

Explanation:

I think this is question from quantisation of charge and could be solved by using formula i.e.,

Q= ne where Q is charge in Si unit

(coulomb)

and n is an integer

and e is charge of one electron..

So given, we required positive charge of 8×10^-7 C

So we have to find value of n which will give us the answer that how many electrons will have given amount amount of charge...

8×10^-7 = n( 1.6 ×10^-19) .........

Where 1.6×10^-19 is charge on one electron( we are not taking -ve sign of electron charge here only magnitude is used because -ve sign only tells that electron has -ve charge on it.

so by here we get

n= 8×10^-7/ 1.6×10^-19

and thus

n= 5×10^12

Please mark it as brainliest if it is right..

I hope it will help. If i am wrong please suggest and also feel free to ask anything related to solution.....