Chapter 18, Problem 23PE

To determine

(a)

The electric field at the center of the triangular configuration of charges.

Answer to Problem 23PE

  $4336\frac{\text{N}}{\text{C}},{\text{ 35}}^{\circ }\text{ below the horizontal}$

Explanation of Solution

Given:

  $\begin{array}{l}{q}_a=+2.50\text{ nC }\\ q_b=-8.00\text{ nC }\\ q_c=+1.50\text{ nC}\end{array}$

Point charges located at the corners of an equilateral triangle 25.0 cm on a side.

Formula used:

Total electric field

  $\overrightarrow{E}=\overrightarrow{E_a}+\overrightarrow{E_b}+\overrightarrow{E_c}$

Calculation:

Let O be the centroid of our equilateral triangle CBA. Distance from O to any vertex is $\frac{a\sqrt{3}}{3}$ where $a=25\text{ cm}$ is the side's length. Since charges given have some arbitrary magnitudes, no useful symmetry can be employed so use of a coordinate system is must. Let O be the origin of a Cartesian system with y axis aligned with OB and x axis as shown above. Let $\overrightarrow{E_a},\overrightarrow{E_b}\text{ and }\overrightarrow{E_c}$ be electric field contributions at O of charges $q_a,q_b\text{ and }{q}_c.$ Total field $\overrightarrow{E}$ at Ocan be calculated as follows:

  $\overrightarrow{E}=\overrightarrow{E_a}+\overrightarrow{E_b}+\overrightarrow{E_c}$

Electric field is generally three-dimensional, but since all charges are coplanar with O, z component vanishes, so find x and y components (projections) of

  $\overrightarrow{E}=\left(E_x,E_y,0\right).$

Notice that $\overrightarrow{E_c}$ forms ${60}^{\circ }$ with y axis and $\overrightarrow{E_a}$ forms ${30}^{\circ }$ with x axis. Using elementary trigonometry:

  $E_x=E_a\cos \left({30}^{\circ }\right)-E_c\sin \left({60}^{\circ }\right)$

  $=\frac{\sqrt{3}}{2}\left(E_a-E_c\right)$

  $=\frac{\sqrt{3}}{2}\frac{k}{{\left(\frac{a\sqrt{3}}{3}\right)}^2}\left(q_a-q_c\right)$

  $=\frac{3\sqrt{3}}{2}\frac{k}{a^2}\left(q_a-q_c\right)$

  $=\frac{3\sqrt{3}}{2}\frac{9\cdot {10}^9\frac{{\text{Nm}}^2}{{\text{C}}^2}}{{0.25}^2{\text{ m}}^2}\left(2.5\cdot {10}^{-9}\text{C}-\text{1}\text{.5}\cdot {\text{10}}^{-9}\text{C}\right)$

  $=374.1\frac{\text{N}}{\text{C}}$

Analogously

  $E_y=E_a\sin \left({30}^{\circ }\right)+E_c\cos \left({60}^{\circ }\right)+E_b$

  $=\frac{1}{2}\left(E_a+E_c+2E_b\right)$

  $=\frac{1}{2}\frac{k}{{\left(\frac{a\sqrt{3}}{3}\right)}^2}\left(q_a+q_c+2\left|q_b\right|\right)$

  $=\frac{3}{2}\frac{k}{a^2}\left(q_a+q_c+2\left|q_b\right|\right)$

  $=\frac{3}{2}\frac{9\cdot {10}^9\frac{{\text{Nm}}^2}{{\text{C}}^2}}{{0.25}^2{\text{m}}^2}\left(2.5+1.5+2\cdot 8\right)\cdot {10}^{-9}\text{C}$

  $=4320\frac{\text{N}}{\text{C}}$

  $\overrightarrow{E}=\left(374.1,4320,0\right)\frac{\text{N}}{\text{C}}.$ Its magnitude is

  $\left|\overrightarrow{E}\right|=\sqrt{{374.1}^2+{4320}^2}\frac{\text{N}}{\text{C}}$

  $=4336\frac{\text{N}}{\text{C}}$

It's tilted ${30}^{\circ }+\arctan \left(\frac{374.1}{4320}\right)={35}^{\circ }$ below the horizontal.

Conclusion:

Thus, $4336\frac{\text{N}}{\text{C}},{\text{ 35}}^{\circ }\text{ below the horizontal}$

To determine

(b)

The combination of charges that will give a zero-strength electric field at the center of the triangular configuration.

Answer to Problem 23PE

There is no combination of charges other than $q_a=q_b=q_c$.

Explanation of Solution

Given:

  $\begin{array}{l}{q}_a=+2.50\text{ nC }\\ q_b=-8.00\text{ nC }\\ q_c=+1.50\text{ nC}\end{array}$

Point charges located at the corners of an equilateral triangle $25\text{ cm}$ on a side.

Calculation:

Suppose $\overrightarrow{E}=0$ at O. Note that all components of $\overrightarrow{E}$ have to be 0 along any direction. Consider line through O parallel with BC. Since component of $\overrightarrow{E}$ along this line must be zero, partial contributions of charges $q_b\text{ and }{q}_c$ along this line cancel each other out. Since $q_b\text{ and }{q}_c$ are equally distanced from O, it is only possible if their magnitudes and signs are equal, so $q_b=q_c$. Now, consider line through O parallel with AB. Since component of $\overrightarrow{E}$ along this line must be zero, partial contributions of charges $q_b\text{ and }{q}_a$ along this line cancel each other out. Since $q_b\text{ and }{q}_a$ are equally distanced from O, it is only possible if their magnitudes and signs are equal, so $q_b=q_a$. Thus,it is shown that $\overrightarrow{E}=0$ at O implies $q_a=q_b=q_c,$ so the answer is No.

Conclusion:

There is no combination of charges other than $q_a=q_b=q_c$.