Chapter 16, Problem 100P

To determine

The $x$ and $y$ components of electric field at a point $P$.

Answer to Problem 100P

The $x$ and $y$ components of electric field at a point $P$ is $\underset{\_}{2.89\times {10}^5\text{N}/\text{C}}$ and $\underset{\_}{2.77\times {10}^6\text{N}/\text{C}}$ respectively.

Explanation of Solution

Write the expression to find the electric field on the x axis.

$E_x=E_{1x}+E_{2x}$ (I)

Here, $E_{1x}$ is the electric field along $x$ axis due to charge 1 and $E_{2x}$ is the electric field along $x$ axis due to charge 2.

Write the expression to find the electric field on the y axis.

$E_y=E_{1y}+E_{2y}$ (II)

Here, $E_{1y}$ is the electric field along $y$ axis due to charge 1 and $E_{2y}$ is the electric field along $y$ axis due to charge 2.

The following figures shows the sketch and the triangular components of the electric field at point $P$.

Conclusion:

Substitute $0$ for $E_{1x}$ and $\frac{k{q}_2}{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}\right)$ for $E_{2x}$ in equation (I).

$\begin{array}{c}{E}_x=0+\frac{k{q}_2}{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}\right)\\ =\frac{k{q}_{2}a}{{\left(a^2+b^2\right)}^{3/2}}\end{array}$

Here, $k$ is a constant, $q_2$ is the second charge, $a$ is the distance between two charges and $b$ is the distance between point $P$ and first charge.

Substitute $8.988\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k$, $47.0\times {10}^{-9}\text{C}$ for $q_2$, $0.0340\text{m}$ for $a$ and $0.0140\text{m}$ for $b$ in the above equation.

$\begin{array}{c}{E}_x=\frac{\left(8.988\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(47.0\times {10}^{-9}\text{C}\right)\left(0.0340\text{m}\right)}{{\left({\left(0.0340\text{m}\right)}^2+{\left(0.0140\text{m}\right)}^2\right)}^{3/2}}\\ =2.89\times {10}^5\text{N}/\text{C}\end{array}$

Substitute $\frac{k{q}_1}{b^2}$ for $E_{1y}$ and $-\frac{k{q}_2}{a^2+b^2}\left(\frac{b}{\sqrt{a^2+b^2}}\right)$ for $E_{2y}$ in equation (II).

$\begin{array}{c}{E}_y=\frac{k{q}_1}{b^2}-\frac{k{q}_2}{a^2+b^2}\left(\frac{b}{\sqrt{a^2+b^2}}\right)\\ =k\left(\frac{q_1}{b^2}-\frac{q_{2}b}{{\left(a^2+b^2\right)}^{3/2}}\right)\end{array}$

Here, $k$ is a constant, $q_2$ is the second charge, $a$ is the distance between two charges and $b$ is the distance between point $P$ and first charge.

Substitute $8.988\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k$, $63.0\times {10}^{-9}\text{C}$ for $q_1$, $47.0\times {10}^{-9}\text{C}$ for $q_2$, $0.0340\text{m}$ for $a$ and $0.0140\text{m}$ for $b$ in the above equation.

$\begin{array}{c}{E}_y=\left(8.988\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(\frac{63.0\times {10}^{-9}\text{C}}{{\left(0.0140\text{m}\right)}^2}-\frac{\left(47.0\times {10}^{-9}\text{C}\right)\left(0.0140\text{m}\right)}{{\left({\left(0.0340\text{m}\right)}^2+{\left(0.0140\text{m}\right)}^2\right)}^{3/2}}\right)\\ =2.77\times {10}^6\text{N}/\text{C}\end{array}$

Therefore, the $x$ and $y$ components of electric field at a point $P$ is $\underset{\_}{2.89\times {10}^5\text{N}/\text{C}}$ and $\underset{\_}{2.77\times {10}^6\text{N}/\text{C}}$ respectively.