Question

Um triângulo equilátero de lado L= 3 cm possui três cargas elétricas iguais a: Q1=2uC, Q2= -3 uC e Q3=8uC instaladas nos seus vértices.

Answer 1

Se o triângulo é equilátero, todas as cargas estão a mesma distância da carga q que está no baricentro do triângulo. Sabemos que do baricentro até um dos vértices vale :

$\displaystyle \frac{2.\text h}{3}$

e

$\displaystyle \text h = \frac{\text L\sqrt{3}}{2}$

Portanto a distância de um dos vértices até o baricentro vale :

$\displaystyle \frac{2.\text h}{3}\to\frac{2}{3}\frac{\text L\sqrt{3}}{2} \to \frac{\text L\sqrt 3 }{3}$

Achando as forças :

$\displaystyle \text{F}_{\text{Q}_1,\text q} = \frac{\text k.\text Q_1.\text q }{(\frac{\text L\sqrt{3}}{3})^2} \to \text{F}_{\text {Q}_1,\text q} = \frac{3.\text k.|\text Q_1|.|\text q|}{\text L^2 }$

Analogamente para as outras forças :

$\displaystyle \text F_{\text Q_2,\text q } =\frac{3\text k|\text Q_2|.|\text q| }{\text L^2} \\\\\\ \displaystyle \text F_{\text Q_3,\text q } =\frac{3\text k|\text Q_3|.|\text q| }{\text L^2}$

Traçando no baricentro uma paralela à base podemos decompor os vetores de Força, sendo ficando apontado para direita os valores :

$\displaystyle \text F_{\text Q_2,\text q }.\text{cos}(30^\circ)+\displaystyle \text F_{\text Q_3,\text q }.\text{cos}(30^\circ)$

E o vetor força  $\displaystyle \text{F}_{\text {Q}_1,\text q}$ apontando para cima perpendicular aos vetores para direita.

Com isso o vetor resultante será :

$\displaystyle \text{Fr}^2= (\text{F}_{\text {Q}_1,\text q})^2+[\text{F}_{\text {Q}_2,\text q}.\text{cos}(30^\circ) + \text{F}_{\text {Q}_3,\text q}.\text{cos}(30^\circ)]^2 \\\\ \text{Fr}^2= (\text{F}_{\text {Q}_1,\text q})^2 + [\text{cos}(30^\circ)]^2.[\text{F}_{\text {Q}_3,\text q}+\text{F}_{\text {Q}_3,\text q}]^2$

Substituindo os valores respectivos valores :

$\displaystyle \text{Fr}^2=(\frac{3.\text k.|\text Q_1|.|\text q|}{\text L^2})^2+[\text{cos}(30^\circ)]^2.(\frac{3.\text k.|\text Q_2|.|\text q|}{\text L^2} + \frac{3.\text k.|\text Q_3|.|\text q|}{\text L^2})^2$

temos :

$\text Q_1 = 2.10^{-6} \ ; \ \text Q_2 = -3.10^{-6} \ ; \ \text Q_3 = 8.10^{-6}\ \text C \\\\ \ \text q = -4.10^{-6} \ \text C \ ; \\\\ \ \text L = 3\text{cm} = 3.10^{-2} \ \text m$

Ou seja :

*

$\displaystyle (\text {F}_{\text Q_1,\text q })^2=[\frac{3.9.10^9.2.10^{-6}.4.10^{-6}}{(3.10^{-2})^2}]^2 \\\\\\ (\text {F}_{\text Q_1,\text q })^2 = [\frac{3.9.2.410^{-12+9}}{9.10^{-4}}]^2 \\\\\ (\text {F}_{\text Q_1,\text q })^2 = [24.10^{-3+4}}]^2 \\\\ (\text {F}_{\text Q_1,\text q })^2 = 24^2.10^2$

*

$\displaystyle [\text{cos}(30^\circ)]^2 = \frac{3}{4} \\\\\\ (\text{F}_{\text Q_2,\text q}+\text{F}_{\text Q_3,\text q} )^2 =(\frac{\text k.\text Q_2.\text q}{\text L^2}+\frac{\text k.\text Q_3.\text q}{\text L^2})^2\\\\\\ (\frac{3.9.10^9.3.10^{-6}.4.10^{-6}}{(3.10^{-2})^2 }+\frac{3.9.10^9.8.10^{-6}.4.10^{-6}}{(3.10^{-2})^2})^2 \\\\\\ (\frac{36.9.10^{-3}+96.9.10^{-3}}{9.10^{-4}})^2 \\\\\\ (\text{F}_{\text Q_2,\text q}+\text{F}_{\text Q_3,\text q} )^2 = 132^2.10^2$

Substituindo na Força resultante :

$\displaystyle \text{Fr}^2=(\frac{3.\text k.|\text Q_1|.|\text q|}{\text L^2})^2+[\text{cos}(30^\circ)]^2.(\frac{3.\text k.|\text Q_2|.|\text q|}{\text L^2} + \frac{3.\text k.|\text Q_3|.|\text q|}{\text L^2})^2$

$\displaystyle \text{Fr}^2=24^2.10^2+\frac{3}{4}.132^2.10^2 \\\\ \text{Fr}^2 = 10^2[576+13068] \\\\ \text{Fr}^2 = 10^2.13644 \\\\ \text{Fr} = \sqrt{10^2.2^2.3^2.379} \\\\ \huge\boxed{\text{Fr} = 60\sqrt{379} \ \text N \ } \checkmark$

( se eu n errei nada é isso aí mesmo )