Chapter 16, Problem 93P

(a)

To determine

The electric field at point P.

Answer to Problem 93P

Electric field is $2.5\times {10}^7\text{N}/\text{C}$ directed at $24°$ below the- x –axis.

Explanation of Solution

Charge $Q_1$ is $-4.5\text{μC}$, position of $Q_1$ is $x=1.00\text{cm}\text{and y=1}\text{.00}\text{cm}$, Charge $Q_2$ is $6.0\text{μC}$, and the position of $Q_2$ is $x=3.00\text{cm}\text{and y=1}\text{.00}\text{cm}$.

Draw the vector diagram showing the direction of electric fields of charges.

Write the equation for the x- component of net electric field at point P.

$E_x=E_{1x}+E_{2x}$ (I)

Here, $E_x$ is the x- component of net electric field at point P, $E_{1x}$ is the x- component of electric field at point P due to charge $Q_1$, and $E_{2x}$ is the x- component of electric field at point P due to charge $Q_2$,

Write the equation for $E_{1x}$.

$E_{1x}=0$ (II)

Write the equation for $E_{2x}$.

$E_{2x}=-k\frac{\left|Q_2\right|}{r_2^2}\left(\frac{2.00}{\sqrt{{2.00}^2+{3.00}^2}}\right)$ (III)

Here, $k$ is the electrostatic constant and $r_2$ is the distance between $Q_2$ and point P.

Rewrite equation (I) by substituting equations (II) and (III).

$\begin{array}{c}{E}_x=0-\frac{2.00k\left|Q_2\right|}{\sqrt{{2.00}^2+{3.00}^2{r}_2^2}}\\ =-\frac{2.00k\left|Q_2\right|}{\sqrt{13.00}{r}_2^2}\end{array}$ (IV)

Write the equation for the y- component of net electric field at point P.

$E_y=E_{1y}+E_{2y}$ (V)

Here, $E_y$ is the y- component of net electric field at point P, $E_{1y}$ is the y- component of electric field at point P due to charge $Q_1$, and $E_{2y}$ is the y- component of electric field at point P due to charge $Q_2$,

Write the equation for $E_{1y}$.

$E_{1y}=-k\frac{\left|Q_1\right|}{r_1^2}$ (VI)

Here, $r_1$ is the distance between $Q_1$ and point P.

Write the equation for $E_{2y}$.

$E_{2y}=k\frac{\left|Q_2\right|}{r_2^2}\left(\frac{3.00}{\sqrt{{2.00}^2+{3.00}^2}}\right)$ (VII)

Rewrite equation (V) by substituting equations (VI) and (VII).

$\begin{array}{c}{E}_y=-k\frac{\left|Q_1\right|}{r_1^2}+k\frac{\left|Q_2\right|}{r_2^2}\left(\frac{3.00}{\sqrt{{2.00}^2+{3.00}^2}}\right)\\ =-k\frac{\left|Q_1\right|}{r_1^2}+k\frac{3.00\left|Q_2\right|}{\sqrt{13.00}{r}_2^2}\end{array}$ (VIII)

Write the equation to find the magnitude of net electric field.

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

Here, $E$ is the magnitude of electric field.

Rewrite the above equation by substituting equations (IV) and (VIII).

$\begin{array}{c}E=\sqrt{{\left(-\frac{2.00k\left|Q_2\right|}{\sqrt{13.00}{r}_2^2}\right)}^2+{\left(-k\frac{\left|Q_1\right|}{r_1^2}+k\frac{3.00\left|Q_2\right|}{\sqrt{13.00}{r}_2^2}\right)}^2}\\ =k\sqrt{{\left(-\frac{2.00\left|Q_2\right|}{\sqrt{13.00}{r}_2^2}\right)}^2+{\left(-\frac{\left|Q_1\right|}{r_1^2}+\frac{3.00\left|Q_2\right|}{\sqrt{13.00}{r}_2^2}\right)}^2}\end{array}$ (IX)

Write the equation to find the direction of $E$.

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

Here, $\theta$ is the angle made by $\overrightarrow{E}$ with negative x-axis.

Conclusion:

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k$, $6.0\text{μC}$ for $Q_2$, $13\text{cm}$ for $r_2$, $3.0\text{cm}$ for $r_1$, and $-4.5\text{μC}$ for $Q_1$ in equation (IX) to find $E$.

$\begin{array}{c}E=\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\sqrt{\begin{array}{l}{\left(-\frac{\left(2.00\right)\left|6.0\text{μC}\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right|}{\sqrt{13.00}{\left(13\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2}\right)}^2+\\ {\left(\frac{\left|-4.5\text{μC}\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right|}{{\left(3.0\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2}+\frac{\left(3.00\right)\left|6.0\text{μC}\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right|}{\sqrt{13.00}{\left(13\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2}\right)}^2\end{array}}\\ =\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(0.278\times {10}^{-2}\text{C}/{\text{m}}^{\text{2}}\right)\\ =2.5\times {10}^7\text{N}/\text{C}\end{array}$

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k$, $6.0\text{μC}$ for $Q_2$, $13\text{cm}$ for $r_2$, $3.0\text{cm}$ for $r_1$, and $-4.5\text{μC}$ for $Q_1$ in equation (X) to find $\theta$.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{\left(\frac{\left|-4.5\text{μC}\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right|}{{\left(3.0\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2}+\frac{\left(3.00\right)\left|6.0\text{μC}\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right|}{\sqrt{13.00}{\left(13\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2}\right)}{\left(-\frac{\left(2.00\right)\left|6.0\text{μC}\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right|}{\sqrt{13.00}{\left(13\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2}\right)}\right)\\ ={\tan }^{-1}\left(0.445\right)\\ =24°\end{array}$

Conclusion:

Therefore, electric field is $2.5\times {10}^7\text{N}/\text{C}$ directed at $24°$ below the- x –axis.

(b)

To determine

The initial acceleration of particle.

Answer to Problem 93P

The initial acceleration is $1.0\times {10}^4\text{m}/{\text{s}}^{\text{2}}$ directed in $24°$ above the +ve x-axis.

Explanation of Solution

Mass of particle is $5.0\text{g}$ and the charge is $-2.0\text{μC}$.

Write the equation to find the net force on the particle.

$F=qE$

Here, $F$ is the net force and $q$ is the charge.

Write the expression for $F$ from Newton’s law of motion.

$F=ma$

Here, $m$ is the mass f particle and $a$ is the acceleration.

Equate the right hand sides of the above two equations.

$\begin{array}{c}qE=ma\\ a=\frac{qE}{m}\end{array}$

Conclusion:

Substitute $-2.0\text{μC}$ for $q$, $2.5\times {10}^7\text{N}/\text{C}$ for $E$, and $5.0\text{g}$ for $m$ in the above equation to find $a$.

$\begin{array}{c}a=\frac{\left(-2.0\text{μC}\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right)\left(2.5\times {10}^7\text{N}/\text{C}\right)}{5.0\text{g}\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)}\\ =1.0\times {10}^{-4}\text{m}/{\text{s}}^{\text{2}}\end{array}$

The direction of acceleration of a negative charge will be opposite to the direction of electric field.

Therefore, the initial acceleration is $1.0\times {10}^4\text{m}/{\text{s}}^{\text{2}}$ directed in $24°$ above the +ve x-axis.