[tex]\text{Given that,}\\\\\text{Electric field strength,}~ E = 81 ~NC^{-1} \\ \\ \text{Distance,}~ d= 49mm = 49 \times 10^{-3} ~m\\ \\\text{Coulomb's constant,}~ k = 9\times 10^9 ~Nm^2 C^{-2}\\\\ \\E= k\dfrac{q}{d^2} \\\\\implies q=\dfrac{Ed^2}{k} = \dfrac{81 \times \left(49 \times 10^{-3}\right)^2}{9\times 10^9} = 2.1609\times 10^{-11} ~C\\\\\\\text{The charge of the particle is}~2.1609\times 10^{-11} ~C[/tex]
