When${10}^{19}$ electrons are removed from a neutral metal plate, the electric charge on it is

According to Quantization of charge

$\text{Q}=\pm \text{ne}$

$\text{Q}=-\text{ne}$ (excess of (-) charges)

$\text{Q}=+\text{ne}$ (remove of (-) charges

$\text{n}={10}^{19},\text{e}=1.6\times {10}^{-19}\text{C}$

$\text{Q}=+\text{ne}$                                  

$\text{=}{\text{+10}}^{19}\times \left(1.6\times {10}^{-19}\right)\text{C}$

$\text{=}\text{}\text{+1}{\text{.6×10}}^{19}\times {10}^{-19}\text{C}$

$\text{=}\text{+1}{\text{.6×10}}^{19-19}\text{C}$

$\text{=}\text{+1}{\text{.6×10}}^0\text{C}$

$\text{=}\text{+1}\text{.6×1C}$

$\text{=}\text{+1}\text{.6C}$