Chapter 21, Problem 21.41E

Three negative point charges lie along a line as shown in Fig. E21.41. Find the magnitude and direction of the electric field this combination of charges produces at point P, which lies 6.00 cm from the −2.00 μC charge measured perpendicular to the line connecting the three charges.

21.41.     IDENTIFY: $\text{E}=\text{k}\frac{\left|\text{q}\right|}{{\text{r}}^2}$. The net field is the vector sum of the fields due to each charge.

SET UP: The electric field of a negative charge is directed toward the charge. Label the charges q₁, q₂, and q₃, as shown in Figure 21.41a. This figure also shows additional distances and angles. The electric fields at point P are shown in Figure 21.41b. This figure also shows the xy-coordinates we will use and the x- and y-components of the fields ${\stackrel{\to }{E}}_1$, ${\stackrel{\to }{E}}_2$, and ${\stackrel{\to }{E}}_3$.

EXECUTE:

$\begin{array}{l}{\text{E}}_1={\text{E}}_3=\left(8.99\times {10}^{9}N·m^2/C^2\right)\frac{5.00\times {10}^{-6}C}{{(0.100m)}^2}=4.49\times {10}^{6}N/C.\\ {\text{E}}_2=(8.99\times {10}^{9}N·m^2/C^2)\frac{2.00\times {10}^{-6}C}{{(0.0600m)}^2}=4.99\times {10}^{6}N/C\end{array}$

E_y = E_1y + E_2y + E_3y = 0 and E_x = E_1x + E_2x + E_3x = E₂ + 2E₁cos53.1° = 1.04 × 10⁷ N/C.

E = 1.04 × 10⁷ N/C, toward the −2.00 μC charge.

EVALUATE: The x-components of the fields of all three charges are in the same direction.

Figure 21.41