Question

. Point charges q1 = q2 = 4.0 × 10−6 C are fixed on the x-axis at x = −3.0 m and x = 3.0 m. What charge q must be placed at the origin so that the electric field vanishes at x = 0, y = 3.0 m?

Answer 1

Answer:

$q_3=-2.83\cdot 10^{-6}\ c$

Explanation:

Electric Field

The total electric field is the sum of all the individual electric fields produced by each charge. Please refer to the image below. The charges q1 and q2 produce the electric fields shown in the point (0,3). Since both charges are equal in magnitude and sign and are at the same distance from the point, the horizontal component of their electric fields cancel each other, only the vertical component remains and it's directed upwards.

The third charge q3 must be negative, so the total electric field is zero. Let's first compute the value of the angle $\theta$ by analyzing the distances in the right triangle formed by the charges q1, q2, and the point (0,3)

$\displaystyle tan\theta=\frac{3}{3}=1$

It follows that $\theta=45^o$

The distance from q1 (or q2) to the point (0,3) is computed by

$d^2=3^2+3^2=18\ m^2$

Let's compute the magnitude of the electric of q1 or q2:

$\displaystyle E=\frac{Kq_1}{d^2}=\frac{Kq_2}{d^2}$

$\displaystyle E=\frac{9\cdot 10^9\times 4\cdot 10^{-6}}{18}$

$E=2000\ N/c$

The vertical component of both fields is given by

$Ey=2000cos45^o=1414.21$

The total field produced by q1 and q2 is twice that quantity

$E_{yt}=2828.42\ N/c$

That should be the value of E3 to make the total field vanish (equal to 0)

$\displaystyle E_3=\frac{Kq_3}{d_3^2}=2828.42$

Solving for q3

$\displaystyle q_3=\frac{2828.42\cdot d_3^2}{K}=\frac{2828.42\cdot 3^2}{9\cdot 10^9}$

$\boxed{ q_3=-2.83\cdot 10^{-6}\ c}$

We had already established the sign of q3 must be negative