Question

alguém me ajuda nessa atividade estou tendo bastante dificuldade​

Answer 1

$q_1=8\ \mu C,\ q_2=-10\ \mu C,\ d=20\ cm=2\times10^{-1}$

a)

$F=k_e\dfrac{q_1q_2}{d^2}\approx(9\times10^9)\dfrac{8\times10^{-6}(-10\times10^{-6})}{(2\times10^{-1})^2}=\\\\ =-\dfrac{720\times10^{-3}}{4\times10^{-2}}=-180\times10^{-1}\ \therefore\ \boxed{F=-18\ N}$

b)

Após o contato, a carga das duas esferas será igual, e terá o valor de:

$q_{1,2}=\dfrac{q_1+q_2}{2}=\dfrac{8\times10^{-6}+(-10)\times10^{-6}}{2}=\boxed{-1\times10^{-6}\ C}$

$d_{1,2}=5\ cm=5\times10^{-2}\ m\\\\ F_{1,2}=k_e\dfrac{q_{1,2}q_{1,2}}{d_{1,2}^2}=(9\times10^9)\dfrac{(-1\times10^{-6})(-1\times10^{-6})}{(5\times10^{-2})^2}=\\\\ =\dfrac{9\times10^{-3}}{25\times10^{-4}}=0.36\times10^{1}\ \therefore\ \boxed{F_{1,2}=3.6\ N}$