Two metal plates having a potential difference of 800V are 2 cm...

$V = 800V$

$E = \dfrac{V}{d} = \dfrac{800}{2 \times {10}^{- 2}}Vm^{- 1}$

$= 4 \times {10}^{4}Vm^{- 1}$

At equilibrium electrostatic force must be equal to force due to weight.

$qE = mg$

$⇒ q = \dfrac{mg}{E}$

$= \dfrac{1. 96 \times {10}^{- 15} \times 9. 8}{4 \times {10}^{4}}$ coulomb

$= \dfrac{1. 96 \times {10}^{- 15} \times 9. 8}{4 \times {10}^{4} \times 1. 6 \times {10}^{- 19}}$ elementary charges

$= 3e$