November 15, 2019

Solution Set 7

Physics 208-02 – Introductory Physics II
Solution Set 7
Nazareth College
Department of Chemistry & Biochemistry
Robert F Szalapski, PhD – Adjunct Lecturer
www.CallMeDrRob.com
Spring 2012

§ Chapter 16, Problem 13

Given an equilateral triangle with an 11.0\;\mu\mathrm{C} charge at each vertex and with side lengths of 15.0\;\mathrm{cm}. Compute the magnitude and direction of the force on each charge. See Figure 1.

Figure 1.

Problem 13 (a).

First of all, notice that each pair of charges will exert equal and opposite charges on each aligned with the sides of the triangle as shown. This is a total of six forces, and all will have the same magnitude due to the same charges at the same separation, but directions are different. Note also that each angle of the triangle is 60^{0}, and hence the angle between the pair of forces at each vertex will also have an angle of 60^{0}.

Figure 2.

Problem 13 (b).

We really want to take advantage of symmetry, and the easiest force to consider if the net force at the top of triangle as shown in Figure 2. Let’s compute the magnitude of each force, and then let’s compute the magnitude of the x and y components as indicated.

\displaystyle F \displaystyle= \displaystyle\left(9.0\times 10^{9}\frac{\mathrm{N\, m}^{2}}{\mathrm{C}^{2}}\right)\frac{\left(11.0\times 10^{{-6}}\mathrm{C}\right)\left(11.0\times 10^{{-6}}\mathrm{C}\right)}{\left(0.150\;\mathrm{m}\right)^{2}}
\displaystyle= \displaystyle 48.4\;\mathrm{N}
\displaystyle F_{x} \displaystyle= \displaystyle F\sin 30^{0}
\displaystyle= \displaystyle 24.2\;\mathrm{N}
\displaystyle F_{y} \displaystyle= \displaystyle F\cos 30^{0}
\displaystyle= \displaystyle 41.\underline{9}2\;\mathrm{N}

Notice that the x component is opposite to the angle, hence the sine function is used for it; the y component is adjacent to the angle, hence the cosine is used for it.
Taking the signs of the components from how the vectors are drawn in the figure,

\displaystyle F_{{1,x}} \displaystyle= \displaystyle 24.2\;\mathrm{N}
\displaystyle F_{{1,y}} \displaystyle= \displaystyle 41.\underline{9}2\;\mathrm{N}
\displaystyle F_{{2,x}} \displaystyle= \displaystyle-24.2\;\mathrm{N}
\displaystyle F_{{2,y}} \displaystyle= \displaystyle 41.\underline{9}2\;\mathrm{N}
\displaystyle\left(F_{1}+F_{2}\right)_{x} \displaystyle= \displaystyle 0
\displaystyle\left(F_{1}+F_{2}\right)_{y} \displaystyle= \displaystyle 83.8\;\mathrm{N}

The resultant is shown in Figure 3 with a magnitude of 83.8\;\mathrm{N} straight up as drawn.

Figure 3.

Problem 13 (c).

By symmetry the other resultant forces acting on the other two charges will have the same magnitude with a direction outward from the triangle center and 30^{0} below the x axis as shown. In terms of absolute angles measure counter-clockwise from the +x axis the net forces are at 90^{0}, 210^{0} and 330^{0}.

§ Chapter 16, Problem 14

Given four +6.00\;\mathrm{mC} charges at the four vertices of a square with 0.100\;\mathrm{m} side lengths, compute the net force on each charge. See figure 4.

Figure 4.

Problem 14.

Notice that there will be three forces on each charge as indicated in the figure. Two of the forces will be along the sides of the square and equal in magnitude to each other where the third will be along the diagonal. Compute the length of the diagonal using Pythagorean theorem as indicated in the drawing. Computing the magnitudes of the forces,

\displaystyle F_{1}=F_{3} \displaystyle= \displaystyle\left(9.0\times 10^{9}\frac{\mathrm{N\, m}^{2}}{\mathrm{C}^{2}}\right)\frac{\left(6.00\times 10^{{-3}}\mathrm{C}\right)\left(6.00\times 10^{{-3}}\mathrm{C}\right)}{\left(0.100\;\mathrm{m}\right)^{2}}
\displaystyle= \displaystyle 3.24\times 10^{7}\;\mathrm{N}
\displaystyle F_{2} \displaystyle= \displaystyle\left(9.0\times 10^{9}\frac{\mathrm{N\, m}^{2}}{\mathrm{C}^{2}}\right)\frac{\left(6.00\times 10^{{-3}}\mathrm{C}\right)\left(6.00\times 10^{{-3}}\mathrm{C}\right)}{\left(\sqrt{2}\;(0.100\;\mathrm{m})\right)^{2}}
\displaystyle= \displaystyle 1.62\times 10^{7}\;\mathrm{N}

Working in components

\displaystyle F_{{1,x}} \displaystyle= \displaystyle 3.24\times 10^{7}\;\mathrm{N}
\displaystyle F_{{1,y}} \displaystyle= \displaystyle 0\;\mathrm{N}
\displaystyle F_{{2,x}} \displaystyle= \displaystyle 1.62\times 10^{7}\;\mathrm{N}\cos 45^{0}
\displaystyle= \displaystyle 1.15\times 10^{7}\;\mathrm{N}
\displaystyle F_{{2,y}} \displaystyle= \displaystyle 1.62\times 10^{7}\;\mathrm{N}\sin 45^{0}
\displaystyle= \displaystyle 1.15\times 10^{7}\;\mathrm{N}
\displaystyle F_{{3,x}} \displaystyle= \displaystyle 0\;\mathrm{N}
\displaystyle F_{{3,y}} \displaystyle= \displaystyle 3.24\times 10^{7}\;\mathrm{N}

To compute the resultant force,

\displaystyle F_{x} \displaystyle= \displaystyle F_{{1,x}}+F_{{2,x}}+F_{{3,x}}
\displaystyle= \displaystyle 3.24\times 10^{7}\;\mathrm{N}+0\;\mathrm{N}+1.15\times 10^{7}\;\mathrm{N}
\displaystyle= \displaystyle 4.4\times 10^{7}\mathrm{N}
\displaystyle F_{y} \displaystyle= \displaystyle F_{{1,y}}+F_{{2,y}}+F_{{3,y}}
\displaystyle= \displaystyle 0\;\mathrm{N}+3.24\times 10^{7}\;\mathrm{N}+1.15\times 10^{7}\;\mathrm{N}
\displaystyle= \displaystyle 4.4\times 10^{7}\mathrm{N}
\displaystyle F \displaystyle= \displaystyle\sqrt{F_{x}^{2}+F_{y}^{2}}
\displaystyle= \displaystyle 6.2\times 10^{7}\mathrm{N}

The resultant forces on each of the four charges will have this same magnitude, all will be aligned along the diagonals and outward from the center of the square. Measured counter-clockwise from the +x axis they will be at 45^{0}, 135^{0}, 225^{0} and 315^{0}.

§ Chapter 16, Problem 15

Given two +6.00\;\mathrm{mC} charges and two -6.00\;\mathrm{mC} charges, alternating +-+-, at the four vertices of a square with 0.100\;\mathrm{m} side lengths, compute the net force on each charge. See figure 5.

Figure 5.

Problem 15.

For this problem we can use the results of Problem 14 noting that the forces aligned with the sides of the square are now attractive and hence have changed direction. Otherwise the individual forces have the same values as in Problem 14. We need to modify the sign of some vector components and recalculate the resultant.

\displaystyle F_{{1,x}} \displaystyle= \displaystyle-3.24\times 10^{7}\;\mathrm{N}
\displaystyle F_{{1,y}} \displaystyle= \displaystyle 0\;\mathrm{N}
\displaystyle F_{{2,x}} \displaystyle= \displaystyle 1.62\times 10^{7}\;\mathrm{N}\cos 45^{0}
\displaystyle= \displaystyle 1.15\times 10^{7}\;\mathrm{N}
\displaystyle F_{{2,y}} \displaystyle= \displaystyle 1.62\times 10^{7}\;\mathrm{N}\sin 45^{0}
\displaystyle= \displaystyle 1.15\times 10^{7}\;\mathrm{N}
\displaystyle F_{{3,x}} \displaystyle= \displaystyle 0\;\mathrm{N}
\displaystyle F_{{3,y}} \displaystyle= \displaystyle-3.24\times 10^{7}\;\mathrm{N}

Notice that there will be three forces on each charge as indicated in the figure. Two of the forces will be along the sides of the square and equal in magnitude to each other where the third will be along the diagonal. Compute the length of the diagonal using Pythagorean theorem as indicated in the drawing. Computing the magnitudes of the forces,

\displaystyle F_{1}=F_{3} \displaystyle= \displaystyle\left(9.0\times 10^{9}\frac{\mathrm{N\, m}^{2}}{\mathrm{C}^{2}}\right)\frac{\left(6.00\times 10^{{-3}}\mathrm{C}\right)\left(6.00\times 10^{{-3}}\mathrm{C}\right)}{\left(0.100\;\mathrm{m}\right)^{2}}
\displaystyle= \displaystyle 3.24\times 10^{7}\;\mathrm{N}
\displaystyle F_{2} \displaystyle= \displaystyle\left(9.0\times 10^{9}\frac{\mathrm{N\, m}^{2}}{\mathrm{C}^{2}}\right)\frac{\left(6.00\times 10^{{-3}}\mathrm{C}\right)\left(6.00\times 10^{{-3}}\mathrm{C}\right)}{\left(\sqrt{2}\;(0.100\;\mathrm{m})\right)^{2}}
\displaystyle= \displaystyle 1.62\times 10^{7}\;\mathrm{N}

Working in components

\displaystyle F_{{1,x}} \displaystyle= \displaystyle 3.24\times 10^{7}\;\mathrm{N}
\displaystyle F_{{1,y}} \displaystyle= \displaystyle 0\;\mathrm{N}
\displaystyle F_{{2,x}} \displaystyle= \displaystyle 1.62\times 10^{7}\;\mathrm{N}\cos 45^{0}
\displaystyle= \displaystyle 1.15\times 10^{7}\;\mathrm{N}
\displaystyle F_{{2,y}} \displaystyle= \displaystyle 1.62\times 10^{7}\;\mathrm{N}\sin 45^{0}
\displaystyle= \displaystyle 1.15\times 10^{7}\;\mathrm{N}
\displaystyle F_{{3,x}} \displaystyle= \displaystyle 0\;\mathrm{N}
\displaystyle F_{{3,y}} \displaystyle= \displaystyle 3.24\times 10^{7}\;\mathrm{N}

To compute the resultant force,

\displaystyle F_{x} \displaystyle= \displaystyle F_{{1,x}}+F_{{2,x}}+F_{{3,x}}
\displaystyle= \displaystyle 3.24\times 10^{7}\;\mathrm{N}+0\;\mathrm{N}+1.15\times 10^{7}\;\mathrm{N}
\displaystyle= \displaystyle 4.4\times 10^{7}\mathrm{N}
\displaystyle F_{y} \displaystyle= \displaystyle F_{{1,y}}+F_{{2,y}}+F_{{3,y}}
\displaystyle= \displaystyle 0\;\mathrm{N}+3.24\times 10^{7}\;\mathrm{N}+1.15\times 10^{7}\;\mathrm{N}
\displaystyle= \displaystyle 4.4\times 10^{7}\mathrm{N}
\displaystyle F \displaystyle= \displaystyle\sqrt{F_{x}^{2}+F_{y}^{2}}
\displaystyle= \displaystyle 6.2\times 10^{7}\mathrm{N}

The resultant forces on each of the four charges will have this same magnitude, all will be aligned along the diagonals and outward from the center of the square. Measured counter-clockwise from the +x axis they will be at 45^{0}, 135^{0}, 225^{0} and 315^{0}.
To compute the resultant force,

\displaystyle F_{x} \displaystyle= \displaystyle F_{{1,x}}+F_{{2,x}}+F_{{3,x}}
\displaystyle= \displaystyle-3.24\times 10^{7}\;\mathrm{N}+0\;\mathrm{N}+1.15\times 10^{7}\;\mathrm{N}
\displaystyle= \displaystyle-2.09\times 10^{7}\mathrm{N}
\displaystyle F_{y} \displaystyle= \displaystyle F_{{1,y}}+F_{{2,y}}+F_{{3,y}}
\displaystyle= \displaystyle 0\;\mathrm{N}-3.24\times 10^{7}\;\mathrm{N}+1.15\times 10^{7}\;\mathrm{N}
\displaystyle= \displaystyle-2.09\times 10^{7}\mathrm{N}
\displaystyle F \displaystyle= \displaystyle\sqrt{F_{x}^{2}+F_{y}^{2}}
\displaystyle= \displaystyle 2.96\times 10^{7}\mathrm{N}

By symmetry all four forces are equal in magnitude and directed from the charge towards the center of the square.

§ Chapter 16, Problem 23

What are the magnitude and direction of the electric force on an electron in a uniform electric field of strength 2360\;\mathrm{N}/\mathrm{C} due east?

The charge of the electron is

\displaystyle q \displaystyle= \displaystyle-e
\displaystyle= \displaystyle-1.60\times 10^{{-19}}\mathrm{C}

The magnitude of the force is given by

\displaystyle F \displaystyle= \displaystyle qE
\displaystyle= \displaystyle\left(1.60\times 10^{{-19}}\mathrm{C}\right)\left(2360\frac{\mathrm{N}}{\mathrm{C}}\right)
\displaystyle= \displaystyle 3.78\times 10^{{-16}}\mathrm{N}

The negative sign on the charge indicates that the force and the electric field are in opposite directions, hence the direction is due west.

§ Chapter 16, Problem 24

A proton is released in a uniform electric field, and it experiences a force of 3.75\times 10^{{-14}}\mathrm{N} toward the south. What are the magnitude and direction of the electric field?

The charge of the proton is

\displaystyle q \displaystyle= \displaystyle+e
\displaystyle= \displaystyle+1.60\times 10^{{-19}}\mathrm{C}

The magnitude of the electric is given by

\displaystyle E \displaystyle= \displaystyle\frac{F}{q}
\displaystyle= \displaystyle\frac{3.75\times 10^{{-14}}\mathrm{N}}{1.60\times 10^{{-19}}\mathrm{C}}
\displaystyle= \displaystyle 2.34\times 10^{{5}}\frac{\mathrm{N}}{\mathrm{C}}

The positive sign on the charge indicates that the force and the electric field are in the same direction, hence the direction is south.

§ Chapter 16, Problem 25

A downward force of 8.4\;\mathrm{N} is is exerted on a -8.8\;\mu\mathrm{C} charge. What are the magnitude and direction of the electric field at this point?

The negative sign on the charge indicate that the electric field and the force are in opposite directions, hence the field is upward. Its magnitude is

\displaystyle E \displaystyle= \displaystyle\frac{F}{q}
\displaystyle= \displaystyle\frac{8.4\;\mathrm{N}}{8.8\times 10^{{-6}}\mathrm{C}}
\displaystyle= \displaystyle 9.5\times 10^{{5}}\frac{\mathrm{N}}{\mathrm{C}}

§ Chapter 16, Problem 26

What are the magnitude and direction of the electric field 20.0\;\mathrm{cm} directly above an isolated 33.0\times 10^{{-6}}\mathrm{C} charge?

The charge is positive, hence it acts as a source, and field lines will be directed radially outward from it. Directly above the field lines will be straight up. At the distance indicated the magnitude is

\displaystyle E \displaystyle= \displaystyle k\frac{q}{r^{2}}
\displaystyle= \displaystyle\left(9.0\times 10^{9}\frac{\mathrm{N\, m}^{2}}{\mathrm{C}^{2}}\right)\frac{33.0\times 10^{{-6}}\mathrm{C}}{(0.200\;\mathrm{m})^{2}}
\displaystyle= \displaystyle 7.4\times 10^{{6}}\frac{\mathrm{N}}{\mathrm{C}}

§ Chapter 16, Problem 28

What is the magnitude and direction at the midpoint between a -8.0\;\mu\mathrm{C} charge and a +7.0\;\mu\mathrm{C} charge separated by 8.0\;\mathrm{cm}?

Figure 6.

Problem 28.

See Figure 6. The positive charge acts as a source, and the negative charge acts as a sink. The evaluation point is 4.0\;\mathrm{m} from either. The fields add constructively with the resultant pointing towards the negative charge with a magnitude

\displaystyle E \displaystyle= \displaystyle k\frac{q_{+}}{r_{+}^{2}}+k\frac{q_{-}}{r_{-}^{2}}
\displaystyle= \displaystyle\left(9.0\times 10^{9}\frac{\mathrm{N\, m}^{2}}{\mathrm{C}^{2}}\right)\frac{7.0\times 10^{{-6}}\mathrm{C}}{(0.040\;\mathrm{m})^{2}}+\left(9.0\times 10^{9}\frac{\mathrm{N\, m}^{2}}{\mathrm{C}^{2}}\right)\frac{8.0\times 10^{{-6}}\mathrm{C}}{(0.040\;\mathrm{m})^{2}}
\displaystyle= \displaystyle 8.4\times 10^{{7}}\frac{\mathrm{N}}{\mathrm{C}}

§ Chapter 16, Problem 34

Calculate the electric field at one corner or a square if the other three corners are occupied by 2.25\times 10^{{-6}}\mathrm{C} charges. See Figure 7.

Figure 7.

Problem 34.

Notice that there will be contributions to the total electric field from each each of three charges as indicated in the figure. Two of the contributions will be along the sides of the square and equal in magnitude to each other where the third will be along the diagonal. Compute the length of the diagonal using Pythagorean theorem as indicated in the drawing. Computing the magnitudes of the forces,

\displaystyle E_{1}=E_{3} \displaystyle= \displaystyle\left(9.0\times 10^{9}\frac{\mathrm{N\, m}^{2}}{\mathrm{C}^{2}}\right)\frac{\left(2.25\times 10^{{-6}}\mathrm{C}\right)}{\left(1.00\;\mathrm{m}\right)^{2}}
\displaystyle= \displaystyle 2.0\underline{2}5\times 10^{4}\frac{\mathrm{N}}{C}
\displaystyle E_{2} \displaystyle= \displaystyle\left(9.0\times 10^{9}\frac{\mathrm{N\, m}^{2}}{\mathrm{C}^{2}}\right)\frac{2.25\times 10^{{-6}}\mathrm{C}}{\left(\sqrt{2}\;(0.100\;\mathrm{m})\right)^{2}}
\displaystyle= \displaystyle 1.0\underline{1}2\times 10^{4}\frac{\mathrm{N}}{C}

Working in components

\displaystyle E_{{1,x}} \displaystyle= \displaystyle 2.0\underline{2}5\times 10^{4}\frac{\mathrm{N}}{C}
\displaystyle E_{{1,y}} \displaystyle= \displaystyle 0\frac{\mathrm{N}}{C}
\displaystyle E_{{2,x}} \displaystyle= \displaystyle 1.0\underline{1}2\times 10^{4}\frac{\mathrm{N}}{C}\cos 45^{0}
\displaystyle= \displaystyle 0.716\times 10^{4}\frac{\mathrm{N}}{C}
\displaystyle E_{{2,y}} \displaystyle= \displaystyle 1.0\underline{1}2\times 10^{4}\frac{\mathrm{N}}{C}\sin 45^{0}
\displaystyle= \displaystyle 0.716\times 10^{4}\frac{\mathrm{N}}{C}
\displaystyle E_{{3,x}} \displaystyle= \displaystyle 0\frac{\mathrm{N}}{C}
\displaystyle E_{{3,y}} \displaystyle= \displaystyle 2.0\underline{2}5\times 10^{4}\frac{\mathrm{N}}{C}

To compute the resultant force,

\displaystyle E_{x} \displaystyle= \displaystyle E_{{1,x}}+E_{{2,x}}+E_{{3,x}}
\displaystyle= \displaystyle 2.0\underline{2}5\times 10^{4}\frac{\mathrm{N}}{C}+0\frac{\mathrm{N}}{C}+0.716\times 10^{4}\frac{\mathrm{N}}{C}
\displaystyle= \displaystyle 2.74\times 10^{4}\frac{\mathrm{N}}{C}
\displaystyle E_{y} \displaystyle= \displaystyle E_{{1,y}}+E_{{2,y}}+E_{{3,y}}
\displaystyle= \displaystyle 0\frac{\mathrm{N}}{C}+2.0\underline{2}5\times 10^{4}\frac{\mathrm{N}}{C}+0.716\times 10^{4}\frac{\mathrm{N}}{C}
\displaystyle= \displaystyle 2.74\times 10^{4}\frac{\mathrm{N}}{C}
\displaystyle E \displaystyle= \displaystyle\sqrt{E_{x}^{2}+E_{y}^{2}}
\displaystyle= \displaystyle 3.88\times 10^{4}\frac{\mathrm{N}}{C}

counter-clockwise at 45^{0} from the +x axis which is outward from the center along the diagonal.

 

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