Questions: No triângulo retângulo isósceles xyz, em que xz=yz=3.0 cm, foram colocadas uma carga elétrica pontual Q4= no vértice x e uma carga elétrica pontual Qv=+B nC A intensidade do campo elétrico resultante em z, devido às cargas já citadas, é Dados o meio e o vácuo e a constante eletrostática do vácuo é k0=9 * 10^9 N * m^2 / C^2 a) 2 * 10^3 N / C. b) 6 * 10^3 N / C c) 8 * 10^4 N / C d) 10^4 N / C e) 10^5 N / C.

Solution Steps

• Charge at point X: \( q_1 = +16 \, \text{nC} \)
• Charge at point Y: \( q_2 = -8 \, \text{nC} \)
• Distance between X and Z: \( r_{XZ} = 3.0 \, \text{cm} = 0.03 \, \text{m} \)
• Distance between Y and Z: \( r_{YZ} = 3.0 \, \text{cm} = 0.03 \, \text{m} \)

Using Coulomb's law for the electric field: \[ E = \frac{k \cdot |q|}{r^2} \] where \( k = 9 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \).

For charge \( q_1 \) at X: \[ E_{XZ} = \frac{9 \times 10^9 \cdot 16 \times 10^{-9}}{(0.03)^2} \] \[ E_{XZ} = \frac{144 \times 10^0}{0.0009} \] \[ E_{XZ} = 1.6 \times 10^5 \, \text{N/C} \]

For charge \( q_2 \) at Y: \[ E_{YZ} = \frac{9 \times 10^9 \cdot 8 \times 10^{-9}}{(0.03)^2} \] \[ E_{YZ} = \frac{72 \times 10^0}{0.0009} \] \[ E_{YZ} = 8 \times 10^4 \, \text{N/C} \]

• The electric field due to \( q_1 \) (positive charge) at Z points away from X.
• The electric field due to \( q_2 \) (negative charge) at Z points towards Y.

Since the triangle is isosceles and the distances are equal, the resultant electric field will be along the vertical axis (Z-axis).

Final Answer