Question

Two 8.00 µC charged particles are located on the x axis. One is at x = 1.00 m, and the other is at x = -1.00 m.

(a) Determine the electric field on the y axis at y = 0.800 m.

N/C i   + N/C j

(b) Calculate the electric force on a -3.00 µC charge placed on the y axis at y = 0.800 m.

N i + N j

Answer 1

a)

The electric field on the \(y\) axis at \(y=0.800\) is

$$ E=\frac{k q}{x^{2}+y^{2}}=\frac{\left(9 \times 10^{9}\right)\left(8.00 \times 10^{-6}\right)}{(1.00)^{2}+(0.800)^{2}}=43902 \mathrm{~N} / \mathrm{C} $$

Angle of each electric field with \(\mathrm{x}\) axis is

$$ \theta=\tan ^{-1}\left(\frac{0.800}{1.00}\right)=38.66^{\circ} $$

The electric field on the \(y\) axis at \(y=0.800 \mathrm{~m}\) along \(\mathrm{x}\) direction is

$$ E_{x}=E \cos \theta-E \cos \theta=0 $$

The electric field on the \(\mathrm{y}\) axis at \(\mathrm{y}=0.800 \mathrm{~m}\) al ong \(\mathrm{y}\) direction is

$$ E_{y}=2 E \sin \theta=2(43902 \mathrm{~N} / \mathrm{C}) \sin 38.66^{\circ}=54850.7 \mathrm{~N} / \mathrm{C} $$

The net electric field on the \(y\) axis at \(y=0.800 \mathrm{~m}\) is

$$ E=(54850.7 \mathrm{~N} / \mathrm{C}) \mathbf{j} $$

b) The electric force on a \(-3.00 \mu \mathrm{C}\) charge placed on the \(y\) axis at \(y=0.800 \mathrm{~m}\) is

$$ \begin{aligned} F=& q E=\left(-3.00 \times 10^{-6} \mathrm{C}\right)((54850.7 \mathrm{~N} / \mathrm{C}) \mathbf{j}) \\ &=-(0.165 \mathrm{~N}) \mathbf{j} \end{aligned} $$