Question

Three point charges are aligned along the x-axis as shown in the figure below. Find the electric field at the position x = +2.4 m, y = 0.

Answer 1

The charges, \(q_{1}=-4.0 \times 10^{-9} \mathrm{C}\)

$$ \begin{array}{l} q_{2}=5.0 \times 10^{-9} \mathrm{C} \\ q_{3}=3.0 \times 10^{-9} \mathrm{C} \end{array} $$

Let the point be \(p(2.4 \mathrm{~m}, 0)\)

The di stances of charges from the point, \(P\) are

$$ \begin{array}{l} r_{1}=2.9 \mathrm{~m} \\ r_{2}=2.4 \mathrm{~m} \\ r_{3}=1.6 \mathrm{~m} \end{array} $$

The electric field at the point,p is given by,

$$ \begin{aligned} E &=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{q_{1}}{r_{1}^{2}}+\frac{q_{2}}{r_{2}^{2}}+\frac{q_{3}}{r_{3}^{2}}\right) \\ &=9 \times 10^{9}\left(\frac{\left(-4.0 \times 10^{-9}\right)}{(2.9)^{2}}+\frac{\left(5.0 \times 10^{-9}\right)}{(2.4)^{2}}+\frac{\left(3.0 \times 10^{-9}\right)}{(1.6)^{2}}\right) \\ &=9(-0.476+0.868+1.72) \\ &=19.01 \mathrm{~N} / \mathrm{C} \end{aligned} $$

The direction is along the positive \(X\) -axis.