Question

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Answer 1

Dados:

$Q=5\times10^{-6}\ C$

$d=4\ m$

$k_0\approx9\times10^9\ Nm^2C^{-2}$

Resolução:

$\boxed{E=\dfrac{k_0Q}{d}}$

$E=\dfrac{(9\times10^9)(5\times10^{-6})}{4^2}=\dfrac{45\times10^{3}}{16}=2.8125\times10^3\ \therefore$

$\boxed{E=2.8125\times10^3\ N/C}$