What are the strength and direction of the electric field at the position indicated by the dot in FIGURE P$23.35$? Give your answer (a) in component form and (b) as a magnitude and angle measured cw or ccw (specify which) from the positive $x$-axis.

The electric field is,

$E=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}$

permittivity of free space is ${\epsilon }_0$.

In the diagram below, two positive charges and one negative charge are located in the three corners of a rectangle.

The electric field generated by the positive charge at the dot is directed away from the positive charge. The electric field generated by the negative charge at the dot is directed towards the negative charge.

a.

Calculate the component form of the electric field strength at the place given by the dot in the diagram.

The angle formed by the vertical and the diagonal is,

$\theta ={\tan }^{-1}\left(\frac{2.0\mathrm{m}}{4.0\mathrm{m}}\right)$

$=26.56°$

The component form of the resultant electric field at the dot is as follows:

$E_r=E_x\hat{i}+E_y\hat{j}$

From figure,

$E_x=E_1-E_3\cos \theta$

$E_y=E_3\sin \theta -E_2$

a.

$x$-component of the net electric field at the dot.

The magnitude of the electric field due to the charge one is

$E_1=\frac{1}{4\pi {\epsilon }_0}\frac{q_1}{r_1^2}$

The magnitude of the electric field due to the charge two is,

$E_2=\frac{1}{4\pi {\epsilon }_0}\frac{q_2}{r_2^2}$

The magnitude of the electric field due to the charge three is,

role="math" localid="1651413236226" $E_3=\frac{1}{4\pi {\epsilon }_0}\frac{q_3}{r_3^2}$

The value,

$r_3=\sqrt{(2.0\mathrm{cm}{)}^2+(4.0\mathrm{cm}{)}^2}$

$=4.47\mathrm{cm}$

$E_x=\frac{1}{4\pi {\epsilon }_0}\frac{q_1}{r_1^2}-\frac{1}{4\pi {\epsilon }_0}\frac{q_3}{r_3^2}\cos \theta$

So,

${\mathrm{E}}_{\mathrm{x}}=\left(9\times {10}^9{\mathrm{C}}^2/\mathrm{N}·{\mathrm{m}}^2\right)\frac{(5.0\mathrm{nC})}{(2.0\mathrm{cm}{)}^2}-\left(9\times {10}^9{\mathrm{C}}^2/\mathrm{N}·{\mathrm{m}}^2\right)\frac{(5.0\mathrm{nC})}{(4.47\mathrm{cm}{)}^2}\left(\cos 26.56°\right)$

$=\left(9\times {10}^9{\mathrm{C}}^2/\mathrm{N}·{\mathrm{m}}^2\right)\left(\frac{\left(5.0\mathrm{nC}\left(\frac{{10}^{-9}\mathrm{C}}{1\mathrm{nC}}\right)\right)}{{\left(2.0\mathrm{cm}\left(\frac{1.0\mathrm{m}}{100\mathrm{cm}}\right)\right)}^2}-\frac{\left(5.0\mathrm{nC}\left(\frac{{10}^{-9}\mathrm{C}}{1\mathrm{nC}}\right)\right)}{{\left(4.47\mathrm{cm}\left(\frac{1.0\mathrm{m}}{100\mathrm{cm}}\right)\right)}^2}\left(\cos 26.56°\right)\right)$

$=1.0\times {10}^5\mathrm{N}/\mathrm{C}$

For $y$component,

$E_y=\frac{1}{4\pi {\epsilon }_0}\frac{q_3}{r_3^2}\sin \theta -\frac{1}{4\pi {\epsilon }_0}\frac{q_2}{r_2^2}$

So,

$E_y=\left(9\times {10}^9{\mathrm{C}}^2/\mathrm{N}·{\mathrm{m}}^2\right)\left(\frac{(5.0\mathrm{nC})}{(4.47\mathrm{cm}{)}^2}(\sin 26.56)-\frac{(10.0\mathrm{nC})}{(4.0\mathrm{cm}{)}^2}\right)$

$=\left(9\times {10}^9{\mathrm{C}}^2/\mathrm{N}·{\mathrm{m}}^2\right)\left(\frac{\left(5.0\mathrm{nC}\left(\frac{{10}^{-9}\mathrm{C}}{1\mathrm{nC}}\right)\right)}{{\left(4.47\mathrm{cm}\left(\frac{1.0\mathrm{m}}{100\mathrm{cm}}\right)\right)}^2}(\sin 26.56)-\frac{\left(10.0\mathrm{nC}\left(\frac{{10}^{-9}\mathrm{C}}{1\mathrm{nC}}\right)\right)}{{\left(4.0\mathrm{cm}\left(\frac{1.0\mathrm{m}}{100\mathrm{cm}}\right)\right)}^2}\right)$

$=-3.6\times {10}^4\mathrm{N}/\mathrm{C}$

So electric dot is,

$E_r=\left(1.0\times {10}^5\mathrm{N}/\mathrm{C}\right)\hat{i}+\left(-3.6\times {10}^4\mathrm{N}/\mathrm{C}\right)\hat{j}$

b.

Magnitude of the electric field at the dot is given by the expression,

$E=\sqrt{E_x^2+E_y^2}$

Substitute all values,

$E=\sqrt{{\left(1.0\times {10}^5\mathrm{N}/\mathrm{C}\right)}^2+{\left(-3.6\times {10}^4\mathrm{N}/\mathrm{C}\right)}^2}$

$=1.1\times {10}^5\mathrm{N}/\mathrm{C}$

angle of electric field is,

${\theta }^{\text{'}}={\tan }^{-1}\left(\frac{E_y}{E_x}\right)$

${\theta }^{\text{'}}={\tan }^{-1}\left(\frac{-\left(3.6\times {10}^4\mathrm{N}/\mathrm{C}\right)}{\left(1.0\times {10}^5\mathrm{N}/\mathrm{C}\right)}\right)$

$=-19.4°$

Here, negative sign indicates counter clockwise direction from the positive$x$-axis.