91Ó°ÊÓ

First, find the distance between each charge and the point where the electric field should be zero. Let the charges be \(q_1\), \(q_2\), \(q_3\), and their positions are given by \((0, 0)\), \((8 m, 0)\), and \((4 m, 0)\). The point of interest is \((4.0 m, 3.0 m)\). For the charge \(q_1\), Distance \(r_{1}=\sqrt{(4-0)^2+(3-0)^2}=\sqrt{16+9}=5.0m\) For the charge \(q_2\), Distance \(r_{2}=\sqrt{(4-8)^2+(3-0)^2}=\sqrt{16+9}=5.0m\) For the charge \(q_3\), Distance \(r_{3}=\sqrt{(4-4)^2+(3-0)^2}=\sqrt{0+9}=3.0m\)

Now, compute the electric field created by each charge at the point of interest. The electric field \(\textbf{E}\) caused by a charge \(q\) at a distance \(r\) is given by \(\textbf{E}=\dfrac{q}{4\pi\epsilon_0r^2}\hat{r}\), where \(\epsilon_0\) is the vacuum permittivity. We will compute the horizontal and vertical components of the electric field for each charge. For \(q_1\), Horizontal component (\(E_{1x}\))=\(\dfrac{q}{4\pi\epsilon_05^2}\cdot\dfrac{4}{5}\) Vertical component (\(E_{1y}\))=\(\dfrac{q}{4\pi\epsilon_05^2}\cdot\dfrac{3}{5}\) For \(q_2\), Horizontal component (\(E_{2x}\))=\(\dfrac{q}{4\pi\epsilon_05^2}\cdot\dfrac{-4}{5}\) Vertical component (\(E_{2y}\))=\(\dfrac{q}{4\pi\epsilon_05^2}\cdot\dfrac{3}{5}\) For \(q_3\), Horizontal component (\(E_{3x}\))=\(0\) Vertical component (\(E_{3y}\))=\(\dfrac{q_3}{4\pi\epsilon_03^2}\cdot\dfrac{3}{3}\)

We want the net electric field to be zero at the point of interest. So, we will add the horizontal and vertical components of the electric fields produced by each charge and set the total electric field equal to zero. Horizontal components of the electric field: \(E_{1x} + E_{2x} + E_{3x} = \dfrac{q}{4\pi\epsilon_05^2}\cdot\dfrac{4}{5} + \dfrac{-q}{4\pi\epsilon_05^2}\cdot\dfrac{4}{5} + 0 = 0\) Vertical components of the electric field: \(E_{1y} + E_{2y} + E_{3y} = \dfrac{q}{4\pi\epsilon_05^2}\cdot\dfrac{3}{5} + \dfrac{q}{4\pi\epsilon_05^2}\cdot\dfrac{3}{5} + \dfrac{q_3}{4\pi\epsilon_03^2}\cdot\dfrac{3}{3} = 0\)

Now, solve the equation for the vertical components of the electric field to find the value of charge \(q_3\). \(\dfrac{q}{4\pi\epsilon_05^2}\cdot\dfrac{3}{5} + \dfrac{q}{4\pi\epsilon_05^2}\cdot\dfrac{3}{5} + \dfrac{q_3}{4\pi\epsilon_03^2}\cdot\dfrac{3}{3} = 0\) \(\dfrac{2q}{5}\cdot\dfrac{1}{25\epsilon_0}+\dfrac{q_3}{9\epsilon_0}=0\) \(q_3=-\dfrac{2q}{9}\left(\dfrac{1}{25}\right)\left(\dfrac{1}{\frac{1}{9}}\right)\) \(q_3=-\dfrac{18}{25}q\) The charge \(q_3\) must be equal to \(-\dfrac{18}{25}q\) to make the electric field at the point \((4.0 m, 3.0 m)\) zero.