Question

When ${10}^{19}$electrons are removed from a neutral metal plate through some process, the charge on it becomes _____?

Answer 1

Step 1: Given that

We remove ${10}^{19}$ electrons from a metal plate, so the number of electrons $\mathrm{n}={10}^{19}$.

Each electron has a charge $\mathrm{e}=1.6\times {10}^{-19}\mathrm{C}$

Step 2: Formula used

From the quantization of charge,

$\mathrm{q}=\pm \mathrm{ne}$

Step 3: Calculation

The total charge removed from the meta can be given by,

$\begin{array}{rcl}\mathrm{q}& =& -\mathrm{ne}\\ & =& -({10}^{19}\times 1.6\times {10}^{-19})\\ & =& -1.6\mathrm{C}\end{array}$

Now applying the conservation of electric charge, when q amount of charge is removed from a metal plate, then an equal and opposite charge will develop on that metal plate, to conserve the total charge in this process.

So,$+1.6\mathrm{C}$ charge will develop on the metal plate after the removal of ${10}^{19}$ electrons.