Chapter 16, Problem 38P

To determine

To Find: The direction of electric field at A and B .

Answer to Problem 38P

The electric field is $4.6\times {10}^6\text{N}/\text{C}$ at point A pointing upwards.

The electric field is $8\times {10}^6\text{N}/\text{C}$ at point B with an angle $15.8°$ above the horizontal.

Explanation of Solution

Given:

Magnitude of charge, $Q=7.0\mu \text{C}$

  $\begin{array}{l}=7.0\mu \text{C}\left(\frac{{10}^{-6}\text{C}}{1\mu C}\right)\\ =7.0\times {10}^{-6}\text{C}\end{array}$

  

Horizontal distance from the charge to midpoint E is $CE=10.0\text{cm}$

  $\begin{array}{l}=10.0\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\\ =0.1\text{m}\end{array}$

Vertical distance from the center of the charges to point A is $AE=5.0\text{cm}$

  $\begin{array}{l}=5.0\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\\ =0.05\text{m}\end{array}$

From Pythagorean Theorem the distance between the left side charge to point $\text{A}$ is

  $\begin{array}{l}d=\sqrt{C{E}^2+A{E}^2}\\ =\sqrt{\left(0.05{\text{m}}^2\right)+{(0.1\text{m})}^2}\\ =0.11\text{m}\end{array}$

From the right-angle triangle $\text{ACE}$

  $\begin{array}{l}\tan \theta =\frac{\text{AE}}{\text{CE}}\\ \begin{array}{cc}& =\frac{5.0\text{cm}}{10.0\text{cm}}\\ \theta & ={\tan }^{-1}(0.5)\\ & ={26.6}^{\circ }\end{array}\end{array}$

Formula used:

Electric field due to charge q at a distance r away from it can be expressed as follows:

  $E=\frac{kq}{r^2}$

Here, k is the Coulomb’s constant.

Calculation:

Electric field at point A: since, both the charges are at same distance from point $\text{A},$ then the horizontal components of electric filed at point A due to both charges will cancel each other. The vertical components of electric field give the net electric field at point $\text{A}$ .

Electric filed due to both charges can be expressed as

  $\begin{array}{l}{E}_A=\frac{kQ}{d^2}\sin \theta +\frac{kQ}{d^2}\sin \theta \\ E_A=2k\frac{Q}{d^2}\sin \theta \end{array}$

On substituting given numerical in the above equation, the electric field at point $\text{A}$ can be expressed as

  $\begin{array}{l}{E}_{\text{A}}=\left(\frac{2\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(7.0\times {10}^{-6}\text{C}\right)}{{(0.111\text{m})}^2}\right)\sin 26.6°\\ =4.6\times {10}^6\text{N}/\text{C}\end{array}$

Electric field at point B:

Point $B$ is not symmetrically placed between the charges; hence one must calculate both horizontal and vertical components.

From the triangle BCD the angle made by left side charge at point B is

  $\begin{array}{c}\tan \theta =\frac{BD}{CD}\\ \tan \theta =\frac{5.0\text{cm}}{5.0\text{cm}}\\ \theta ={\tan }^{-1}\left(1.0\right)\\ \theta =45°\end{array}$

From the triangle BDF, the angle made by right side charge at point B is

  $\begin{array}{c}\tan \theta =\frac{BD}{DF}\\ \tan \theta =\frac{5.0\text{cm}}{15.0\text{cm}}\\ \theta ={\tan }^{-1}\left(0.333\right)\\ \theta =18.43°\end{array}$

The distance between left side charges 1 to point B is

  $\begin{array}{l}BC=\sqrt{B{D}^2+C{D}^2}\\ BC=\sqrt{{(0.05\text{m})}^2+{(0.05\text{m})}^2}\\ BC=0.0707\text{m}\end{array}$

From the triangle $\Delta BDF$ ,the distance between the right-side charge to point B is

  $\begin{array}{l}BF=\sqrt{{(BD)}^2+{(DF)}^2}\\ =\sqrt{{(0.05\text{m})}^2+{(0.15\text{m})}^2}\\ =0.1581\text{m}\end{array}$

The net horizontal electric field due to left side charge at point B can be expressed as

  $\begin{array}{l}{E}_x={\overrightarrow{E}}_{\text{lef t}}+{\overrightarrow{E}}_{\text{right }}\\ =\frac{kQ}{B{C}^2}\cos {\theta }_{\text{left}}-\frac{kQ}{B{F}^2}\cos {\theta }_{\text{right }}\\ =kQ\left(\frac{\cos {\theta }_{\text{left }}-\cos {\theta }_{\text{right }}}{B{C}^2-B{F}^2}\right)\\ =\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(7.0\times {10}^{-6}\text{C}\right)\left(\frac{\cos 45°-\cos 18.43°}{{(0.0707\text{m})}^2-{(0.1581\text{m})}^2}\right)\\ =7.6\times {10}^6\text{N}/\text{C}\end{array}$

The net vertical electric field due to right side charge at point B can be expressed as

  $\begin{array}{l}{E}_y={\overrightarrow{E}}_{\text{lcft}}+{\overrightarrow{E}}_{\text{right }}\\ =\frac{kQ}{B{C}^2}\sin {\theta }_{\text{left }}+\frac{kQ}{B{F}^2}\sin {\theta }_{\text{right }}\\ =kQ\left(\frac{\sin {\theta }_{\text{left }}+\sin {\theta }_{\text{right }}}{B{C}^2+B{F}^2}\right)\\ =\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(7.0\times {10}^{-6}\text{C}\right)\left(\frac{\sin 45°+\sin 18.43°}{{(0.0707\text{m})}^2+{(0.1581\text{m})}^2}\right)\\ =2.15\times {10}^6\text{N}/\text{C}\end{array}$

The net electric field at point B can be expressed as

  $\begin{array}{l}{E}_B=\sqrt{E_x^2+E_y^2}\\ =\sqrt{{\left(7.6\times {10}^6\text{N}/\text{C}\right)}^2+{\left(2.15\times {10}^6\text{N}/\text{C}\right)}^2}\\ \approx 8\times {10}^6\text{N}/\text{C}\end{array}$

The direction of electric field at point B can be expressed as

  $\begin{array}{l}\tan \theta =\frac{E_y}{E_x}\\ \theta ={\tan }^{-1}\left(\frac{E_y}{E_x}\right)\\ ={\tan }^{-1}\left(\frac{2.15\times {10}^6\text{N}/\text{C}}{7.6\times {10}^6\text{N}/\text{C}}\right)\\ =15.8°\end{array}$

Conclusion:

The electric field is $8\times {10}^6\text{N}/\text{C}$ at point B with an angle $15.8°$ above the horizontal.