Chapter 24, Problem 32SP

Charges of +2.0, +3.0, and $-8.0\mu \text{C}$ are placed in air at the vertices of an equilateral triangle of side 10 cm. Calculate the magnitude of the force acting on the $-8.0\mu \text{C}$ charge due to the other two charges.

To determine

The magnitude of force acting on the $-8.0\mu \text{C}$ charge due to the other two charges if $+2.0,+3.0,\text{ and }-8.0\mu \text{C}$ charges are placed at corner of the triangles.

Answer to Problem 32SP

Solution:

$31.38\text{ N}$

Explanation of Solution

Given Data:

The change on point A of the triangle is $+2.0\mu \text{C}$.

The change on point B of the triangle is $+3.0\mu \text{C}$.

The change on point C of the triangle is $-8.0\mu \text{C}$.

The length of each side of the triangle is $10\text{ cm}$.

Formula Used:

Write the expression for the electrostatic force:

$F=\frac{k\left|q_1\right|\left|q_2\right|}{r^2}$

Here, $F$ is the electrostatic force the between two charges, $k$ is the Coulomb’s constant, $q_1$ is the magnitude of charge on charge 1, $q_2$ is the magnitude of charge on charge 2, and $r$ is the physical distance between the two charges.

The expression for resultant electrostatic force acting on one charge (test charge) due to presence of two other charges is given as

$F_r=\sqrt{F_1^2+F_2^2+2F_1{F}_2\mathrm{Cos}\theta }$

Here, Here, $F_r$ is the resultant electrostatic force acting one charge due to presence of two other charges, $F_1$ is the electrostatic force acting on test charge due to charge 1, $F_2$ is the electrostatic force acting on test charge due to charge 2, and $\theta$ is the angle between directions of force $F_1$ and $F_2$.

Explanation:

Assume an equilateral triangle ABC having each side equal to $10\text{ cm}$. One charge is placed on each of the three vertices such that charge on point A is $+2.0\mu \text{C}$, point B is $+3.0\mu \text{C}$, and test charge on point C is $-8.0\mu \text{C}$.

The electrostatic force exerted by charge placed at A on charged placed at C is $F_{AC}$ and the electrostatic force exerted by charge placed at B on charged placed at C is $F_{BC}$.

Here, first calculate the electrostatic force exerted by charge placed at A on charge placed at B. The expression for the electrostatic force is given as

$F_{AB}=\frac{k{q}_A{q}_B}{r_{AB}^2}$

Here, $F_{AB}$ is the electrostatic force exerted by charge placed at point A on charged placed at point B, $k$ is the Coulomb’s constant, $q_A$ is the magnitude of charge on point A, $q_B$ is the magnitude of charge on point B, and $r_{AB}$ is the physical distance between charges A and B.

Substitute $9\times {10}^9\text{ N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $+2.0\mu \text{C}$ for $q_A$, $-8.0\mu \text{C}$ for $q_B$, and $10\text{ cm}$ for $r_{AB}$

$\begin{array}{c}{F}_{AB}=\frac{\left(9\times {10}^9\text{ N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left|\left(+2.0\mu \text{C}\right)\right|\left|\left(-8.0\mu \text{C}\right)\right|}{{\left(10\text{ cm}\right)}^2}\\ =\frac{\left(9\times {10}^9\text{ N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(2\mu \text{C}\right)\left(\frac{{10}^{-6}\text{ C}}{1\mu \text{C}}\right)\left(8.0\mu \text{C}\right)\left(\frac{{10}^{-6}\text{ C}}{1\mu \text{C}}\right)}{{\left(10\text{ cm}\left(\frac{1\text{ m}}{{10}^2\text{ cm}}\right)\right)}^2}\\ =\frac{\left(9\times {10}^9\text{ N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(2\mu \text{C}\right)\left(8\mu \text{C}\right)}{{\left(10\text{ cm}\right)}^2}\\ =14.4\text{ N}\end{array}$

Now, calculate the electrostatic force exerted by charge placed at B on charge placed at C.

The expression for the electrostatic force is given as follows:

$F_{BC}=\frac{k{q}_B{q}_C}{r_{BC}^2}$

Here, $F_{BC}$ is the electrostatic force exerted by charge placed at point B on charged placed at point C, $k$ is the Coulomb’s constant, $q_B$ is the magnitude of charge on point B, $q_C$ is the magnitude of charge on point C, and $r_{BC}$ is the physical distance between charges B and C.

Substitute $9\times {10}^9\text{ N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $+3.0\mu \text{C}$ for $q_B$, $-8.0\mu \text{C}$ for $q_C$, and $10\text{ cm}$ for $r_{BC}$

$\begin{array}{c}{F}_{AB}=\frac{\left(9\times {10}^9\text{ N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left|\left(+3.0\mu \text{C}\right)\right|\left|\left(-8.0\mu \text{C}\right)\right|}{{\left(10\text{ cm}\right)}^2}\\ =\frac{\left(9\times {10}^9\text{ N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(\text{3 }\mu \text{C}\right)\left(\frac{{10}^{-6}\text{ C}}{1\mu \text{C}}\right)\left(8.0\mu \text{C}\right)\left(\frac{{10}^{-6}\text{ C}}{1\mu \text{C}}\right)}{{\left(10\text{ cm}\left(\frac{1\text{ m}}{{10}^2\text{ cm}}\right)\right)}^2}\\ =21.6\text{ N}\end{array}$

Finally, we shall calculate the resultant force acting on charge placed at C due to the forces exerted by charges placed at A and C.

The expression for resultant electrostatic force acting on charge placed at C due to the forces exerted by charges placed at A and C is given as

$F_r=\sqrt{F_{AC}^2+F_{BC}^2+2F_{AC}{F}_{BC}\mathrm{Cos}\theta }$

Here, Here, $F_r$ is the resultant electrostatic force acting one charge due to presence of two other charges, $F_{AC}$ is the electrostatic force exerted by charge placed at point A on charged placed at point C, $F_{BC}$ is the electrostatic force exerted by charge placed at point B on charged placed at point C, and $\theta$ is the angle between directions of force $F_{AC}$ and $F_{BC}$.

Substitute $-\text{14}\text{.4 N}$ for $F_{AC}$, $-\text{21}\text{.6 N}$ for $F_{BC}$, and $\text{60}°$ for $\theta$

$\begin{array}{c}{F}_r=\sqrt{{\left(14.4\text{ N}\right)}^2+{\left(21.6\text{ N}\right)}^2+2\left(14.4\text{ N}\right)\left(21.6\text{ N}\right)\left(\cos 60°\right)}\\ =\sqrt{\left(207.36{\text{ N}}^2\right)+\left(466.56{\text{ N}}^2\right)+\left(622.08{\text{ N}}^2\right)\left(0.5\right)}\\ =\sqrt{984.96}\text{ N}\\ =\text{31}\text{.38 N}\end{array}$

Conclusion:

The magnitude of force acting on the $-8.0\mu \text{C}$ charge due to the other two charges is $31.38\text{ N}$