Chapter 19, Problem 1OQ

A point charge of −4.00 nC is located at (0, 1.00) m. What is the x component of the electric field due to the point charge at (4.00, −2.00) m? (a) 1.15 N/C (b) −0.864 N/C (c) 1.44 N/C (d) −1.15 N/C (e) 0.864 N/C

To determine

The $x$ component of the electric field due to the point charge.

Answer to Problem 1OQ

Option (d) $-1.15\text{N}/\text{C}$.

Explanation of Solution

The point charge of $-4.00\text{ nC}$ is displaced from $\left(0,1.00\right)\text{ m}$ to the point $\left(4.00,-2.00\right)\text{ m}$.

Write the formula to find the magnitude of displacement

    $r=\sqrt{r_x^2+r_y^2}$        (I)

Here, $r$ is the displacement, $r_x$ is the displacement in $x$ direction and $r_y$ is the displacement in $y$ direction.

Write the formula for electric field

    $E=\frac{k_{e}q}{r^2}$        (II)

Here, $E$ is the electric field, $k_e$ is Coulomb constant $k_e=8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ and $q$ is the charge.

Conclusion:

The displacement of the point charge in $x$ direction is

  $\begin{array}{c}{r}_x=\left(x_f-x_i\right)\\ =\left(4.00-0\right)\text{ m}\\ =4.00\text{ m}\end{array}$

The displacement of the point charge in $y$ direction is

    $\begin{array}{c}{r}_y=\left(x_y-x_i\right)\\ =\left(-2.00-1.00\right)\text{ m}\\ =-3.00\text{ m}\end{array}$

Substitute $4.00\text{ m}$ for $r_x$ and $-3.00\text{ m}$ for $r_y$ in equation (I) to find the value of $r$

$\begin{array}{c}r=\sqrt{{\left(4.00\right)}^2+{\left(-3.00\right)}^2}\\ =5.00\text{ m}\end{array}$

The direction of the point charge displaced is

    $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{r_y}{r_x}\right)\\ ={\tan }^{-1}\left(\frac{-3.00}{4.00}\right)\\ =36.9°\end{array}$

The $x$ component of electric field due to the point charge is $E_x=E\cos \theta$.

Substitute equation (II) in the above relation

$E_x=\frac{k_{e}q}{r^2}\cos \theta$

Substitute $36.9°$ for $\theta$, $8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k_e$, $-4.00\times {10}^{-9}\text{ C}$ for $q$ and $5.00\text{ m}$ for $r$ in the above equation to find the value of $E_x$

$\begin{array}{c}{E}_x=\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(-4.00\times {10}^{-9}\text{ C}\right)}{{\left(5.00\text{ m}\right)}^2}\cos 36.9°\\ =-1.15\text{N}/\text{C}\end{array}$

The $x$ component of the electric field due to the point charge at the point $\left(4.00,-2.00\right)\text{ m}$ is $-1.15\text{N}/\text{C}$.