Chapter 16, Problem 51QAP

To determine

The net force exerted on each charge by other charges.

Answer to Problem 51QAP

Net force on charge A is 36.3 N at angle $24.4°$ below negative x -axis. Net force on charge B is 43.45 N at angle $11°$ below the positive x -axis. Net force on charge C is 25.2 N at angle ${67.5}^{°}$ above the negative x -axis.

Explanation of Solution

Given:

Charge A, charge B and charge C are placed in a coordinate system as shown below.

  

Formula used:

From Coulomb's law

  $\begin{array}{l}{F}_{q_1\text{ on }{q}_2}=\frac{k|q_1\Vert q_2|}{r^2}\\ \text{where},\\ r\text{=}\text{seperation}\text{between}\text{charge}\text{1}\text{and}\text{charge}\text{2}\end{array}$

Calculation:

For charge A,

x component of net force on charge A :

  $\begin{array}{l}{\displaystyle \sum F_{\text{net on }A,x}}=F_{B\text{ on }A,x}+F_{C\text{ on }A,x}=F_{B\text{ on }A,x}+0=\frac{k|q_B\Vert q_A|}{r_{AB}^2}\\ =\frac{\left(8.99\times {10}^9\frac{\text{N}\cdot {\text{m}}^2}{{\text{C}}^2}\right)\left|6.00\times {10}^{-6}\text{C}\right|\left|3.00\times {10}^{-6}\text{C}\right|}{{\left(7.00\times {10}^{-2}\text{m}\right)}^2}\\ =33.02\text{N}\end{array}$

Because charges A and B have the same sign, the force on charge A is repulsive and acts in the

negative x direction.

Thus, the net force on charge A in the x -direction is - 33.0 N.

The y component of net force on charge A :

  $\begin{array}{l}{\displaystyle \sum F_{\text{net on }A,y}}=F_{B\text{ on }A,y}+F_{C\text{ on }A,y}=0+F_{C\text{ on }A,y}=\frac{k|q_C\Vert q_A|}{r_{AC}^2}\\ {\displaystyle \sum F_{\text{net on }A,y}}=\frac{\left(8.99\times {10}^9\frac{\text{N}\cdot {\text{m}}^2}{{\text{C}}^2}\right)\left|2.00\times {10}^{-6}\text{C}\right|\left|3.00\times {10}^{-6}\text{C}\right|}{{\left(6.00\times {10}^{-2}\text{m}\right)}^2}\\ {\displaystyle \sum F_{\text{net on }A,y}}=14.98\text{N}\end{array}$

Because charges A and C have the same sign, the force on charge A is repulsive and acts in the

negative y -direction.

Thus, the net force on charge A in the y -direction is -15.0 N

The magnitude of the net force on charge A :

  $\begin{array}{l}{F}_{\text{ net on }A}=\sqrt{{\left(F_{\text{ net on }A,x}\right)}^2+{\left(F_{\text{ net on }A,y}\right)}^2}\\ F_{\text{ net on }A}=\sqrt{{(-33.02\text{N})}^2+{(-14.98\text{N})}^2}\\ F_{\text{ net on }A}=36.3N\end{array}$

The direction of the net force on charge A :

  ${\theta }_A={\tan }^{-1}\left(\frac{-14.98\text{N}}{-33.02\text{N}}\right)={24.4}^{°}$

For charge B,

x component of net force on charge B :

  ${\displaystyle \sum F_{\text{net on }B,x}}=F_{A\text{ on }B,x}+F_{\text{C}\text{on}B,x}=\frac{k\left|q_A\right|\left|q_B\right|}{r_{AB}^2}+\frac{k|q_C\Vert q_B|}{r_{CB}^2}\cos {\theta }_{CB}$

  ${\displaystyle \sum F_{\text{net on }B,x}}=\frac{k|q_A\Vert q_B|}{r_{AB}^2}+\frac{k|q_{\text{C}}\Vert q_B|}{\left(r_{AB}^2+r_{AC}^2\right)}\left(\frac{r_{AB}}{\sqrt{r_{AB}^2+r_{AC}^2}}\right)$

  ${\displaystyle \sum F_{\text{net on }B,x}}=\frac{k|q_A\Vert q_B|}{r_{AB}^2}+\frac{k|q_{\text{C}}\Vert q_B|r_{AB}}{{\left(r_{AB}^2+r_{AC}^2\right)}^{3/2}}$

  ${\displaystyle \sum F_{\text{net on }B,x}}=\frac{\left(8.99\times {10}^9\frac{\text{N}\cdot {\text{m}}^2}{{\text{C}}^2}\right)\left|3.00\times {10}^{-6}\text{C}\right|\left|6.00\times {10}^{-6}\text{C}\right|}{{\left(7.00\times {10}^{-2}\text{m}\right)}^2}$

  $+\frac{\left(8.99\times {10}^9\frac{\text{N}\cdot {\text{m}}^2}{{\text{C}}^2}\right)\left|2.00\times {10}^{-6}\text{C}\right|\left|6.00\times {10}^{-6}\text{C}\right|\left(7.00\times {10}^{-2}\text{m}\right)}{{\left[{\left(7.00\times {10}^{-2}\text{m}\right)}^2+{\left(6.00\times {10}^{-2}\text{m}\right)}^2\right]}^{3/2}}$

  ${\displaystyle \sum F_{\text{net on }B,x}}=42.66\text{N}$

y component of net force on charge B :

  $\begin{array}{l}{\displaystyle \sum F_{\text{net on }B,y}}=F_{A\text{ on }B,y}+F_{\text{C}\text{on}B,y}=0+\frac{k|q_{\text{C}}\Vert q_B|}{r_{BC}^2}\sin {\theta }_{CB}\\ =\frac{k|q_{\text{C}}\Vert q_B|}{r_{AB}^2+r_{AC}^2}\left(\frac{r_{AC}}{\sqrt{r_{AB}^2+r_{AC}^2}}\right)\\ =\frac{k|q_C\Vert q_B|r_{AC}}{{\left(r_{AB}^2+r_{AC}^2\right)}^{3/2}}\\ =\frac{\left(8.99\times {10}^9\frac{\text{N}\cdot {\text{m}}^2}{{\text{C}}^2}\right)\left|2.00\times {10}^{-6}\text{C}\right|\left|6.00\times {10}^{-6}\text{C}\right|\left(6.00\times {10}^{-2}\text{m}\right)}{{\left[{\left(7.00\times {10}^{-2}\text{m}\right)}^2+{\left(6.00\times {10}^{-2}\text{m}\right)}^2\right]}^{3/2}}\\ =8.260\text{N}\end{array}$

Because charges C and B have the same sign, the force on charge B is repulsive and acts in the negative y-direction. Thus, the net force on charge B in the y -direction is -8.26 N

The magnitude of the net force on charge B :

  $\begin{array}{l}{F}_{\text{ net on }B}=\sqrt{{\left(F_{\text{ net on }B,x}\right)}^2+{\left(F_{\text{ net on }B,y}\right)}^2}\\ F_{\text{ net on }B}=\sqrt{{(42.66\text{N})}^2+{(-8.26\text{N})}^2}\\ F_{\text{ net on }B}=43.45\text{N}\end{array}$

The direction of the net force on charge B :

  ${\theta }_B={\tan }^{-1}\left(\frac{-8.260\text{N}}{42.66\text{N}}\right)=-{11.0}^{°}$

For charge C,

x component of net force on charge C :

  $\begin{array}{l}{\displaystyle \sum F_{\text{net on }}}{c}_{,x}=F_{A\text{ on }C,x}+F_{B\text{ on }C,x}=0+\frac{k|q_B\Vert q_C|}{r_{BC}^2}\cos {\theta }_{BC}\\ {\displaystyle \sum F_{\text{net on }}}{c}_{,x}=\frac{k|q_B\Vert q_C|}{r_{AB}^2+r_{AC}^2}\left(\frac{r_{AB}}{\sqrt{r_{AB}^2+r_{AC}^2}}\right)\\ {\displaystyle \sum F_{\text{net on }}}{c}_{,x}=\frac{k|q_B\Vert q_C|r_{AB}}{{\left(r_{AB}^2+r_{AC}^2\right)}^{3/2}}\\ {\displaystyle \sum F_{\text{net on }}}{c}_{,x}=\frac{\left(8.99\times {10}^9\frac{\text{N}\cdot {\text{m}}^2}{{\text{C}}^2}\right)\left|6.00\times {10}^{-6}\text{C}\right|\left|2.00\times {10}^{-6}\text{C}\right|\left(7.00\times {10}^{-2}\text{m}\right)}{{\left[{\left(7.00\times {10}^{-2}\text{m}\right)}^2+{\left(6.00\times {10}^{-2}\text{m}\right)}^2\right]}^{3/2}}\\ {\displaystyle \sum F_{\text{net on }}}{c}_{,x}=9.636\text{N}\text{(direction}\text{of}\text{force}\text{is}\text{towards}\text{negative}\text{x-axis)}\end{array}$

Because charges C and B have the same sign, the force on charge C is repulsive and acts in the

negative x direction. Thus, the net force on charge C in the x-direction is -9.64 N

y component of net force on charge C :

  ${\displaystyle \sum F_{\text{net}\text{on}C,y}}=F_{A\text{on}C,y}+F_{B\text{ on }C,y}=\frac{k|q_A\Vert q_{\text{C}}|}{r_{AC}^2}+\frac{k\left|q_B\right|\left|q_{\text{C}}\right|}{r_{BC}^2}\sin {\theta }_{BC}$

  ${\displaystyle \sum F_{\text{net}\text{on}C,y}}=\frac{k|q_A\Vert q_C|}{r_{AC}^2}+\frac{k|q_B\Vert q_C|}{r_{AB}^2+r_{AC}^2}\left(\frac{r_{AC}}{\sqrt{r_{AB}^2+r_{AC}^2}}\right)$

  ${\displaystyle \sum F_{\text{net}\text{on}C,y}}=\frac{k\left|q_A\right|\left|q_C\right|}{r_{AC}^2}+\frac{k\left|q_B\right|\Vert q_C|r_{AC}}{{\left(r_{AB}^2+r_{AC}^2\right)}^{3/2}}$

  ${\displaystyle \sum F_{\text{net}\text{on}C,y}}=\frac{\left(8.99\times {10}^9\frac{\text{N}\cdot {\text{m}}^2}{{\text{C}}^2}\right)\left|3.00\times {10}^{-6}\text{C}\right|\left|2.00\times {10}^{-6}\text{C}\right|}{{\left(6.00\times {10}^{-2}\text{m}\right)}^2}$

  $+\frac{\left(8.99\times {10}^9\frac{\text{N}\cdot {\text{m}}^2}{{\text{C}}^2}\right)\left|6.00\times {10}^{-6}\text{C}\right|\left|\text{l}2.00\times {10}^{-6}\text{C}\right|\left(6.00\times {10}^{-2}\text{m}\right)}{{\left[{\left(7.00\times {10}^{-2}\text{m}\right)}^2+{\left(6.00\times {10}^{-2}\text{m}\right)}^2\right]}^{3/2}}$

  ${\displaystyle \sum F_{\text{net}\text{on}C,y}}=23.24\text{N}$

The magnitude of the net force on charge C :

  $\begin{array}{l}{F}_{netonC}=\sqrt{{\left(F_{netonC,x}\right)}^2+{\left(F_{netonC,y}\right)}^2}\\ F_{netonC}=\sqrt{{(-9.636N)}^2+{(23.24N)}^2}\\ F_{netonC}=25.2N\end{array}$

The direction of the net force on charge C :

  ${\theta }_{\text{C}}={\tan }^{-1}\left(\frac{23.24\text{N}}{-9.636\text{N}}\right)=-{67.5}^{°}$

Conclusion:

Net force on charge A is 36.3 N at angle $24.4°$ below negative x -axis. Net force on charge B is 43.45 N at angle $11°$ below the positive x -axis. Net force on charge C is 25.2 N at angle ${67.5}^{°}$ above the negative x -axis.