Question

how many electrons must be removed from a piece of metal to give it a positive charge of 1 x 10^-7 C..???​

Answer 1

Answer:

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Using

Quantization Principle of Charges....

q = +-ne..

we get

n = q/e =

$\frac{1 \times {10}^{ - 7} }{1.6 \times {10}^{ - 19} } = 6.25 \times {10}^{11}$

where ....

n = no. of electrons...

q = charge

e = charge on electron = 1.6 x 10^-19.

Answer 2

Given:

• Piece of metal has charge +1 × 10^(-7) C.

To find:

• How many electrons were removed?

Calculation:

• First of all, we can understand that the body lost electrons and hence achieved positive charge.

• Now, the charge of each electron is -1.6 × 10^(-19) C (this is the smallest charge possible).

Now, number of electrons removed :

$\therefore \: no. \: of \: {e}^{ - } = \dfrac{1 \times {10}^{ - 7} }{1.6 \times {10}^{ - 19} }$

$\implies \: no. \: of \: {e}^{ - } = 0.625 \times {10}^{ 19 - 7}$

$\implies \: no. \: of \: {e}^{ - } = 0.625 \times {10}^{ 12}$

$\implies \: no. \: of \: {e}^{ - } = 6.25 \times {10}^{ 11}$

So, number of electrons removed is 6.25 × 10¹¹ .