Find the charge on an iron particle of mass 2.24 mg, if 0.02% of electrons are removed from it.

0.01996 C

As $\frac{\text { mass }}{\text { Atomic wt. }}=\frac{\text { No.of atoms }}{\text {Avogadro No. }}$

∴ No. of atoms = $24 \times 10^{18}$ atoms $=24 \times 10^{18} \times 26$ electrons.

n = No. of electrons removed = $24 \times 10^{18} \times 26 \times \frac{0.01}{100}=1248 \times 10^{14}$ electron

∴ Q = ne = (+ve charge) = 0.01996 C