Question

Determine a intensidade do Campo Elétrico E, de uma carga Q= 8x10^-6C distante a 2 cm. Considere meio vácuo,K= 9x10^9N.M^2/C^2.

Answer 1

$\boxed{F_e_l_e_t = \frac{Kq^2}{d^2}}$

$F_e_l_e_t = \frac{9.10^9(8.10^-^6)^2}{(2.10^-^3)^2}\\\\\\ F_e_l_e_t = \frac{9.10^9(64.10^-^1^2)}{(4.10^-^6})\\\\\\ \boxed{F_e_l_e_t = 144.10^3\ N}$

$F_e_l_e_t = E*q\\\\\ 144.10^3 = 8.10^-^6E\\\\\ E = \frac{144.10^3}{8.10^-^6}\\\\\ \boxed{E = 1,8.10^1^0\ N/C}$

Answer 2

Q=8×10⁻⁶C
d=2 cm⇒2×10⁻²
K=9×10⁹

E= K×Q/d²
E=9×10⁹×8×10⁶/(2×10⁻²)²
E=72×10¹⁵ /4×10⁻⁴
E=18×10¹⁹ N/C