Chapter 19, Problem 68P

Two point charges q_A = −12.0 μC and q_B = 45.0 μC and a third particle with unknown charge q_C are located on the x axis. The particle q_A is at the origin, and q_B is at x = 15.0 cm. The third particle is to be placed so that each particle is in equilibrium under the action of the electric forces exerted by the other two particles. (a) Is this situation possible? If so, is it possible in more than one way? Explain. Find (b) the required location and (c) the magnitude and the sign of the charge of the third particle.

To determine

Whether the situation is possible or not.

Answer to Problem 68P

The situation is possible in two ways.

Explanation of Solution

Given Info: Charge at the origin is $-12\text{ μC}$, and charge at $15\text{ cm}$ distance is $45\text{ μC}$.

When the charges are of the same sign, the force is repulsive and when the charges are of the opposite sign, the force is attractive. In order to achieve equilibrium, the forces due to the charges must be balanced in magnitude as well as in direction.

Thus, it is possible to attain equilibrium with the charge $q_3$ in two ways.

• if a negative charge is placed on the right of charge $+45\text{ μC}$.
• if a positive charge is placed on the left of charge $-12\text{ μC}$.

Conclusion:

Therefore, it is possible to attain equilibrium in two ways.

To determine

The location of the charge.

Answer to Problem 68P

The location of the charge is $16\text{ cm}$ to the left of the charge $-12\text{ μC}$.

Explanation of Solution

Given Info: Charge at the origin is $-12\text{ μC}$, and charge at $15\text{ cm}$ distance is $45\text{ μC}$.

According to Coulomb’s law, the electric field created by a charge $q$ is,

$\overrightarrow{\text{E}}=k_e\frac{q}{r^2}\widehat{r}$

Here,

$E$ is the electric field.

$k_e$ is the Coulomb’s constant.

$q$ is the charge.

$r$ is the distance between two charges.

The figure for the given conditions is shown below.

Figure (1)

Case (1):

If a negative charge is placed on the right of charge $+45\text{ μC}$.

The electric field due to charge $q_1$ is,

$\overrightarrow{{\text{E}}_1}=k_e\frac{q_1}{r_1{}^2}{\widehat{r}}_1$ . (1)

Here,

$E_1$ is the electric field due to charge $q_1$.

$q_1$ is the charge at the origin.

$r_1$ is the distance between $q_1$ and the third charge $q_3$.

The electric field due to charge $q_2$ is,

$\overrightarrow{{\text{E}}_2}=k_e\frac{q_2}{r_2{}^2}{\widehat{r}}_2$ (2)

Here,

$E_2$ is the electric field due to charge $q_2$.

$q_2$ is the charge at the origin.

$r_2$ is the distance between $q_2$ and the third charge $q_3$.

In order to achieve equilibrium both $E_1$ and $E_2$ will be equal.

Equate equation (1) and equation (2),

$\begin{array}{c}{E}_1=E_2\\ k_e\frac{q_1}{r_1{}^2}=k_e\frac{q_2}{r_2{}^2}\\ \frac{q_1}{r_1{}^2}=\frac{q_2}{r_2{}^2}\end{array}$ (3)

Substitute $12.0\text{ μC}$ for $q_1$, $45.0\text{ μC}$ for $q_2$ and $\left(x+15\right)$ for $r_1$ and $x$ for $r_2$.

$\begin{array}{c}\frac{\left(12.0\text{μC}\right)\left(\frac{1\text{C}}{{10}^6\text{μC}}\right)}{{\left(x+15\right)}^2}=\frac{\left(45.0\text{μC}\right)\left(\frac{1\text{C}}{{10}^6\text{μC}}\right)}{x^2}\\ {\left(\frac{x}{x+15}\right)}^2=\frac{15}{4}\\ 4x^2=15\left(x^2+30x+225\right)\\ 11x^2+450x+3375=0\end{array}$

Solve the above quadratic equation for $x$.

$x=-9.89$ and $x=-31$

As both of the above values for $x$ are negative, it is not possible to put the charge $q_3$ on the right side of charge $q_2$.

Case (2):

If a positive charge is placed on the left of charge $-\text{12 μC}$

Substitute $12.0\text{ μC}$ for $q_1$, $45.0\text{ μC}$ for $q_2$ and $y$ for $r_1$ and $\left(y+15\right)$ for $r_2$ in equation (3) to get $y$,

$\begin{array}{c}\frac{\left(12.0\text{ μC}\right)\left(\frac{1\text{ C}}{{10}^6\text{ μC}}\right)}{y^2}=\frac{\left(45.0\text{ μC}\right)\left(\frac{1\text{ C}}{{10}^6\text{ μC}}\right)}{{\left(y+15\right)}^2}\\ {\left(\frac{y+15}{y}\right)}^2=\frac{15}{4}\\ 4\left(y^2+30y+225\right)=15y^2\\ 11y^2-120y-900=0\end{array}$

Solve the above quadratic equation for $y$,

$y=16$ and $y=-5.11$

Neglect the negative value because $y$ can’t be negative.

Thus, the $y$ will be $16\text{ cm}$.

Conclusion:

Therefore, the location of the charge is $16\text{ cm}$ to the left of the charge $-12\text{ μC}$.

To determine

The magnitude and the sign of the charge of the third particle.

Answer to Problem 68P

The magnitude and the sign of the charge of the third particle is $\text{+51}\text{.2 μC}$

Explanation of Solution

Given Info: Charge at the origin is $-12\text{ μC}$, and charge at $15\text{ cm}$ distance is $45\text{ μC}$.

The expression for the electric force according to Coulomb’s law can be given as,

$F=k_e\frac{q_1{q}_2}{r^2}$

Here,

$F$ is the force due to electric field

$k_e$ is the Coulomb’s constant

$q_1$ is the charge on the first particle

$q_2$ is the charge on the second particle

$r$ is the separation between the two charges.

The expression for the electric force due to $q_3$ on $q_1$ can be given as,

$F_3=k_e\frac{q_1{q}_3}{r_3{}^2}$ (3)

Here,

$F_3$ is the electric force due to charge $q_3$

$r_3$ is the separation between the charge  $q_3$ and $q_1$

The expression for the electric force due to $q_1$ on $q_2$ can be given as,

$F_1=k_e\frac{q_1{q}_2}{r_1{}^2}$ (4)

Here,

$F_1$ is the electric force due to charge $q_1$

$r_1$ is the separation between the charge  $q_1$ and $q_2$

As each particle is in equilibrium under the action of electric forces exerted by the other two particles, both $F_1$ and $F_3$ will be equal.

Equate equation (3) and equation (4),

$\begin{array}{c}{F}_1=F_3\\ k_e\frac{q_1{q}_2}{r_1{}^2}=k_e\frac{q_1{q}_3}{r_3{}^2}\\ \frac{q_2}{r_1{}^2}=\frac{q_3}{r_3{}^2}\\ q_3=\frac{q_2{r}_3{}^2}{r_1{}^2}\end{array}$

Substitute $45\text{ μC}$ for $q_2$, $16\text{ cm}$ for $r_3$ and $15\text{ cm}$ for $r_1$ to get $q_3$

$\begin{array}{c}{q}_3=\frac{\left(45\text{ μC}\right)\left(\frac{1\text{ C}}{{10}^6\text{ μC}}\right){\left(\left(16\text{ cm}\right)\left(\frac{1\text{ m}}{100\text{ cm}}\right)\right)}^2}{{\left(\left(15\text{ cm}\right)\left(\frac{1\text{ m}}{100\text{ cm}}\right)\right)}^2}\\ =51.2\times {10}^{-6}\text{ C}\\ =\text{51}\text{.2 μC}\end{array}$

Thus, the charge $q_3$ will be $\text{51}\text{.2 μC}$.

Conclusion:

Therefore, the magnitude and sign of the third charge $q_3$ will be $\text{+51}\text{.2 μC}$.