Chapter 18, Problem 22PE

To determine

(a)

The electric field at the location of $q_a$.

Answer to Problem 22PE

The electric field at the location of $q_a$ is $1.25\cdot {10}^6\frac{\text{N}}{\text{C}}$.

Explanation of Solution

Given:

  $q_b=+10.00\mu \text{C}$

  $q_c=-5.00\mu \text{C}$

Side of an equilateral triangle $\text{25}\text{.0 cm}$

Formula used:

  $E=E_b\cos \phi$

Calculation:

Let $\overrightarrow{E_b}\text{ and }\overrightarrow{E_c}$ be electric field contributions of charges $q_b\text{ and }{q}_c$ at A.

Total electric field $\overrightarrow{E}$ can be calculated as follows:

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

Since angle at A is ${60}^{\circ },$ angle between $\overrightarrow{E_b}\text{ and }\overrightarrow{E_c}{\text{ is 120}}^{\circ }.$ Distances from $q_b\text{ and }{q}_c$ to A are equal, so $E_c$ is half of $E_b$ because $q_c$ is half of $q_b.$

Consider triangle $A{P}_1{P}_2$ in the picture.

Angle at P₁ is ${180}^{\circ }-{120}^{\circ }={60}^{\circ }$ for P₁ and A are adjacent vertices of a parallelogram.

Also, $\left|P_1{P}_2\right|=\frac{1}{2}\left|A{P}_1\right|$ so by the side-angle-side (SAS) postulate we find that it is similar to a right triangle with angles ${60}^{\circ }\text{ and }{30}^{\circ }.$

Therefore, $\measuredangle P_{2}A{P}_1\equiv \phi ={30}^{\circ },\measuredangle P_1{P}_{2}A={90}^{\circ }.$

  $\begin{array}{c}E=E_b\cos \phi \\ =\frac{k{q}_b}{a^2}\frac{\sqrt{3}}{2}\\ =\frac{9\cdot {10}^9\frac{{\text{Nm}}^2}{{\text{C}}^2}10\cdot {10}^{-6}\text{C}}{{\left(0.25\text{m}\right)}^2}\frac{\sqrt{3}}{2}\\ =1.247\cdot {10}^6\frac{\text{N}}{\text{C}}\end{array}$

Conclusion:

Thus, the electric field at the location of $q_a$ is $1.25\cdot {10}^6\frac{\text{N}}{\text{C}}$.

To determine

(b)

The force on $q_a$.

Answer to Problem 22PE

The force on $q_a$ is $1.87\cdot {10}^{-3}\text{ N}$

Explanation of Solution

Given:

  $q_a=+1.50\text{ nC}$

  $q_b=+10.00\mu \text{C}$

  $q_c=-5.00\mu \text{C}$

Side of an equilateral triangle $\text{25}\text{.0 cm}$

Formula used:

  $F_a=E{q}_b$

Calculation:

  $F_a=E{q}_b$

Substitute the values

  $\begin{array}{c}{F}_a=\left(1.247\times {10}^6\frac{\text{N}}{\text{C}}\right)\left(1.5\times {10}^{-9}\text{C}\right)\\ =1.87\times {10}^{-3}\text{ N}\end{array}$

Conclusion:

Thus, the force on $q_a$ is $1.87\times {10}^{-3}\text{ N}$