Question

a current of 20 microampere flows in a wire for 30 seconds find the amount of charge in the number of free electron transfer to the wire

Answer 1

No of electrons (n) = 6.25 x 10¹⁸ x i x t

                               = 6.25 x 10¹⁸ x 20 x 10⁻⁶ x 30

                                = 3750 x 10¹²

amount of charge = n x e

                               = 3750 x 10¹² x 1.6 x 10⁻¹⁹

                                = 6000 x 10⁻⁷Coulombs = 6 x 10⁻⁴ Coulombs

Answer 2

Given :

Current , I = 20 mA .

Time , t = 30 s .

To Find :

The amount of charge and the number of free electron transfer to the wire .

Solution :

We know , charge is given by :

$Q=It\\\\Q=20\times 10^{-3}\times 30\\\\Q=0.6\ C$

So , number of electron ,

$n=\dfrac{Q}{q_e}\\\\n=\dfrac{0.6}{1.6\times 10^{-19}}\\\\n=3.75\times 10^{18}$

Hence , this is the required solution .

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