Chapter 15, Problem 24P

(a) Find the magnitude and direction of the electric field at the position of the 2.00 μC charge in Figure P13.13. (b) How would the electric field at that point be affected if the charge there were doubled? Would the magnitude of the electric force be affected?

To determine

The magnitude and direction of net electric field on $2\text{μC}$.

Answer to Problem 24P

Solution:

The magnitude of net electric field on $2\text{μC}$ is $2.19\times {10}^5\text{N/C}$.

The direction of net electric field on $2\text{μC}$ is ${85.2}^o$ below the x axis.

Explanation of Solution

The force diagram is given by,

In the above diagram,

• $F_1$ is the force due to $q_1$ on $q_2$.
• $F_3$ is the force due to $q_3$ on $q_2$.

Formula to calculate the force due to $q_1$ on $q_2$ is,

$F_1=\frac{k_e{q}_1{q}_2}{a^2}$      (I)

• $k_e$ is the Coulomb constant.
• a is the side of the triangle

Formula to calculate the force due to $q_3$ on $q_2$ is,

$F_3=\frac{k_e{q}_3{q}_2}{a^2}$      (II)

Net force along the x direction is,

$F_x=F_3-F_1\cos {60}^o$      (III)

Net force along the y direction is,

$F_y=-F_1\sin {60}^o$      (IV)

Formula to calculate the magnitude of net electric force is,

$F_R=\sqrt{F_x^2+F_y^2}$      (V)

Substitute Equations (III) and (IV) in (V).

$\begin{array}{c}{F}_R=\sqrt{{\left(\frac{k_e{q}_3{q}_2}{a^2}-\left(\frac{k_e{q}_1{q}_2}{a^2}\right)\cos {60}^o\right)}^2+{\left(-\frac{k_e{q}_1{q}_2}{a^2}\sin {60}^o\right)}^2}\\ =\left(\frac{k_e{q}_2}{a^2}\right)\sqrt{{\left(q_3-q_1\cos {60}^o\right)}^2+{\left(-q_1\sin {60}^o\right)}^2}\end{array}$

Formula to calculate the resultant electric field is,

$E_R=\frac{F_R}{q_2}$

Therefore,

$E=\left(\frac{q_2}{a^2}\right)\sqrt{{\left(q_3-q_1\cos {60}^o\right)}^2+{\left(-q_1\sin {60}^o\right)}^2}$

Substitute $8.99\times {10}^9\text{N}{\text{.m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k_e$, $7.00\text{μC}$ for $q_1$, $2.00\text{μC}$ for $q_2$, $-4.00\text{μC}$ for $q_3$ and 0.500 m for a.

$\begin{array}{c}{E}_R=\left[\frac{\left(8.99\times {10}^9\text{N}{\text{.m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)}{{\left(0.500\text{m}\right)}^2}\right]\sqrt{{\left(\left(4.00\text{μC}\right)-\left(7.00\text{μC}\right)\cos {60}^o\right)}^2+{\left(\left(7.00\text{μC}\right)\sin {60}^o\right)}^2}\\ =\left[\frac{\left(8.99\times {10}^9\text{N}{\text{.m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)}{{\left(0.500\text{m}\right)}^2}\right]\sqrt{{\left(\left(4.00\times {\text{10}}^{-6}\text{C}\right)-\left(7.00\times {\text{10}}^{-6}\text{C}\right)\cos {60}^o\right)}^2+{\left(\left(7.00\times {\text{10}}^{-6}\text{C}\right)\sin {60}^o\right)}^2}\\ =2.19\times {10}^5\text{N/C}\end{array}$

Formula to calculate the direction is,

$\theta ={\tan }^{-1}\left(\frac{F_y}{F_x}\right)$

From Equations (I), (II), (III) and (IV) of section 1,

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-\frac{k_e{q}_1{q}_2}{a^2}\sin {60}^o}{\frac{k_e{q}_3{q}_2}{a^2}-\left(\frac{k_e{q}_1{q}_2}{a^2}\right)\cos {60}^o}\right)\\ ={\tan }^{-1}\left(\frac{-q_1\sin {60}^o}{q_3-q_1\cos {60}^o}\right)\end{array}$

Substitute $7.00\text{μC}$ for $q_1$, $2.00\text{μC}$ for $q_2$ and $-4.00\text{μC}$ for $q_3$.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-\left(7.00\text{μC}\right)\sin {60}^o}{\left(4.00\text{μC}\right)-\left(7.00\text{μC}\right)\cos {60}^o}\right)\\ =-{85.2}^o\end{array}$

Conclusion:

The magnitude of net electric field on $2\text{μC}$ is 0.438 N.

To determine

The electric field if the charge is doubled.

Answer to Problem 24P

Solution: The electric field remains the same.

Explanation of Solution

When the charge is doubled, the electric force also gets doubled. Therefore, the electric field remains the same, since it is equal to the ration of electric force and charge.

Conclusion:

The electric field remains the same