!! Veja a imagem !!
[tex]\displaystyle * \ \underline{\text{Tra{\c c}{\~a}o (T)}}: \\\\ \text{T}_\text x = \text T.\text{sen}(30^\circ) \\\\ \text{T}_\text y = \text T.\text {cos}(30^\circ) \\\\\\ * \ \text{For{\c c}a el{\'e}trica = Fe} \\\\\\\ \underline{\text{Campo el{\'e}trico (E)}}: \\\\ \text E = \frac{\text{Fe}}{|\text q|}[/tex]
Item a)
A tração na vertical equilibrou com o peso da esfera, então :
[tex]\displaystyle \text T_\text y = \text P \\\\ \text T.\text{cos}(30^\circ) = \text{m.g} \\\\ \frac{\text T.\sqrt{3}}{2} = \text{m.g} \\\\ \boxed{\text T = \frac{2\sqrt{3}.\text{m.g}}{3}}[/tex]
A força elétrica equilibrou com a tração na horizontal, logo :
[tex]\displaystyle\text T_\text x = \text{Fe} \\\\ \text{Fe}=\text T.\text{sen}(30^\circ) \\\\ \text{Fe}=\frac{2\sqrt{3}.\text{m.g}}{3}.\frac{1}{2} \\\\\\ \underline{\text{Portanto a for{\c c}a e{\'e}trica que atua na esfera vale}}:\\\\ \text{Fe}= \frac{\sqrt{3}.\text{m.g}}{3}\\\\ *\text m = 25\ \text g = 25.10^{-3} \ \text{kg}\\\\ \huge\boxed{\text{Fe}=\frac{25.10^{(-3)}.\sqrt{3}.\text g}{3} \ \ \text{N } \ }\checkmark[/tex]
( Não sei se é pra aproximar os valores, aí vc substitui as respectivas aproximações da gravidade e da raiz de 3)
item b)
[tex]\displaystyle \text{E}=\frac{\text{Fe}}{|\text q|} \\\\\\ \text E = \frac{25.10^{(-3)}.\sqrt{3}}{3}.\frac{\text g}{5.10^{-6}} \\\\\\ \text E =\frac{5.10^{(-3+6)}.\sqrt{3}.\text g}{3} \\\\\\ \huge\boxed{\text E = \frac{5.10^{3}.\sqrt{3}.\text g}{3} \ \text{N/C}}\checkmark[/tex]
(mesma coisa, vc substitui as respectivas aproximações da gravidade e da raiz 3)
