Question

A long uniformly charged thread (linear charge density A -2.1 C/m) lies along the z axis in the figure.(Figure 1) A small charged sphere (Q =-2.4 C ) is at the point x=0 cm, y = -5.0 cm.

Part A

What is the direction of the electric field at the point x = 7.0 cm, y =7.0 cm? Ethread and EQ represent fields due to the long thread and the charge Q, respectively.

Part B 

What is the magnitude of the electric field at the point x = 7.0 cm, y = 7.0 cm? 

Answer 1

$$ \begin{aligned} &\vec{E}_{Q}=-\frac{2.4 \times 9 \times 10^{9}}{\left(12^{2}+7^{2}\right) / 10^{4}} \frac{7 \hat{\imath}+12 \hat{\jmath}}{\sqrt{193}} \\ &\vec{E}_{Q}=-1.12 \times 10^{12} \cdot \frac{7 \hat{\imath}+12 \hat{\jmath}}{\sqrt{193}} \\ &\vec{E}_{\text {thread }}=\frac{2 K \lambda}{d} \hat{d} \\ &\vec{E}_{\text {thread }}=\frac{2 \times 9 \times 10^{9} \times 2.1}{0.07} \hat{\jmath}=5.4 \times 10^{11} \hat{\jmath} \end{aligned} $$

Thus, the net field is obtained by adding the 2 ,

$$ \vec{E}=-(5.64 \hat{\imath}+4.26 \hat{\jmath}) \times 10^{11} \mathrm{~N} / \mathrm{C} $$

Direction

$$ 180+\tan ^{-1} \frac{4.26}{5.64}=217.06^{\circ} $$

Magnitude

$$ \sqrt{5.64^{2}+4.26^{2}} \times 10^{11}=7.07 \times 10^{11} \mathrm{~N} / \mathrm{C} $$