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PROBLEM 1 Object A has a charge of +2 mC, and object B has a charge of +6 mC. Which statement is true?

Physics

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2 parts
a

8.24×10−8 N8.24 \times 10^{-8} \text{ N}

3.63×10−47 N3.63 \times 10^{-47} \text{ N}

2.27×10392.27 \times 10^{39}

b

9.05×1022 m/s29.05 \times 10^{22} \text{ m/s}^2

2.66 N2.66 \text{ N}

repulsive\text{repulsive}

Step-by-step solution

6 steps

Step 1: Understand the relationship between forces.
According to Newton's Third Law, when two objects exert forces on each other, these forces are equal in magnitude and opposite in direction. This applies to electrostatic forces as well. The force exerted by object A on object B (F⃗AB\vec{F}_{AB}) is equal in magnitude and opposite in direction to the force exerted by object B on object A (F⃗BA\vec{F}_{BA}).

Step 2: Apply Newton's Third Law.
The mathematical representation of this relationship is:

F⃗AB=−F⃗BA\vec{F}_{AB} = -\vec{F}_{BA}

The magnitudes of the charges (+2 mC+2 \text{ mC} and +6 mC+6 \text{ mC}) determine the magnitude of the force, but they do not change this fundamental vector relationship.

The correct statement is F⃗AB=−F⃗BA\boxed{\vec{F}_{AB} = -\vec{F}_{BA}}.

Step 1: State Coulomb's Law for the initial force.
The initial force F1F_1 between two charges q1q_1 and q2q_2 separated by a distance dd is given by Coulomb's Law:

F1=k∣q1q2∣d2F_1 = k \frac{|q_1 q_2|}{d^2}

where kk is Coulomb's constant.

Step 2: Determine the new force when q1q_1 is tripled.
If q1q_1 is tripled, the new charge is q1′=3q1q_1' = 3q_1. The new force F2F_2 will be:

F2=k∣(3q1)q2∣d2F_2 = k \frac{|(3q_1) q_2|}{d^2}

Step 3: Compare the new force to the initial force.
We can rewrite F2F_2 by factoring out the constant 3:

F2=3⋅k∣q1q2∣d2F_2 = 3 \cdot k \frac{|q_1 q_2|}{d^2}

Since F1=k∣q1q2∣d2F_1 = k \frac{|q_1 q_2|}{d^2}, we can substitute F1F_1 into the equation for F2F_2:

F2=3F1F_2 = 3F_1

The new force will be three times the original force.

The new force F2F_2 will be 3F1\boxed{3F_1}.

Given values:
Distance r=5.3×10−11 mr = 5.3 \times 10^{-11} \text{ m}
Elementary charge e=1.602×10−19 Ce = 1.602 \times 10^{-19} \text{ C}
Mass of electron me=9.109×10−31 kgm_e = 9.109 \times 10^{-31} \text{ kg}
Mass of proton mp=1.672×10−27 kgm_p = 1.672 \times 10^{-27} \text{ kg}
Coulomb's constant k=8.987×109 N⋅m2/C2k = 8.987 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2
Gravitational constant G=6.674×10−11 N⋅m2/kg2G = 6.674 \times 10^{-11} \text{ N} \cdot \text{m}^2/\text{kg}^2

Part a· 3 steps

Find the magnitudes of the electric force and the gravitational force, and their ratio.

  1. Calculate the electric force (FeF_e).

    The electric force between the electron (qe=−eq_e = -e) and proton (qp=+eq_p = +e) is given by Coulomb's Law:

    Fe=k∣qeqp∣r2=ke2r2F_e = k \frac{|q_e q_p|}{r^2} = k \frac{e^2}{r^2} Fe=(8.987×109 N⋅m2/C2)(1.602×10−19 C)2(5.3×10−11 m)2F_e = (8.987 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2) \frac{(1.602 \times 10^{-19} \text{ C})^2}{(5.3 \times 10^{-11} \text{ m})^2} Fe=(8.987×109)2.566404×10−382.809×10−21 NF_e = (8.987 \times 10^9) \frac{2.566404 \times 10^{-38}}{2.809 \times 10^{-21}} \text{ N} Fe=8.24×10−8 NF_e = 8.24 \times 10^{-8} \text{ N}

  2. Calculate the gravitational force (FgF_g).

    The gravitational force between the electron and proton is given by Newton's Law of Universal Gravitation:

    Fg=Gmempr2F_g = G \frac{m_e m_p}{r^2} Fg=(6.674×10−11 N⋅m2/kg2)(9.109×10−31 kg)(1.672×10−27 kg)(5.3×10−11 m)2F_g = (6.674 \times 10^{-11} \text{ N} \cdot \text{m}^2/\text{kg}^2) \frac{(9.109 \times 10^{-31} \text{ kg})(1.672 \times 10^{-27} \text{ kg})}{(5.3 \times 10^{-11} \text{ m})^2} Fg=(6.674×10−11)1.523×10−572.809×10−21 NF_g = (6.674 \times 10^{-11}) \frac{1.523 \times 10^{-57}}{2.809 \times 10^{-21}} \text{ N} Fg=3.63×10−47 NF_g = 3.63 \times 10^{-47} \text{ N}

  3. Calculate the ratio of the electric force to the gravitational force (Fe/FgF_e / F_g).

    FeFg=8.24×10−8 N3.63×10−47 N\frac{F_e}{F_g} = \frac{8.24 \times 10^{-8} \text{ N}}{3.63 \times 10^{-47} \text{ N}} FeFg=2.27×1039\frac{F_e}{F_g} = 2.27 \times 10^{39}

    The electric force is 8.24×10−8 N\boxed{8.24 \times 10^{-8} \text{ N}}, the gravitational force is 3.63×10−47 N\boxed{3.63 \times 10^{-47} \text{ N}}, and their ratio is 2.27×1039\boxed{2.27 \times 10^{39}}.

Part b

9.05×1022 m/s29.05 \times 10^{22} \text{ m/s}^2

Compute the acceleration caused by the electric force of the proton on the electron.

  1. Calculate the acceleration of the electron (aea_e).

    Using Newton's Second Law, Fe=meaeF_e = m_e a_e, the acceleration of the electron due to the electric force is:

    ae=Femea_e = \frac{F_e}{m_e} ae=8.24×10−8 N9.109×10−31 kga_e = \frac{8.24 \times 10^{-8} \text{ N}}{9.109 \times 10^{-31} \text{ kg}} ae=9.05×1022 m/s2a_e = 9.05 \times 10^{22} \text{ m/s}^2

    The acceleration caused by the electric force on the electron is 9.05×1022 m/s2\boxed{9.05 \times 10^{22} \text{ m/s}^2}.

    Given values:
    Charge q1=4.3 µC=4.3×10−6 Cq_1 = 4.3 \text{ µC} = 4.3 \times 10^{-6} \text{ C}
    Charge q2=4.3 µC=4.3×10−6 Cq_2 = 4.3 \text{ µC} = 4.3 \times 10^{-6} \text{ C}
    Distance r=0.25 mr = 0.25 \text{ m}
    Coulomb's constant k=8.987×109 N⋅m2/C2k = 8.987 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2

  2. Determine the type of force.

    Since both charges are positive (+4.3 µC+4.3 \text{ µC}), they will repel each other.

  3. Calculate the magnitude of the force using Coulomb's Law.

    F=k∣q1q2∣r2F = k \frac{|q_1 q_2|}{r^2} F=(8.987×109 N⋅m2/C2)(4.3×10−6 C)(4.3×10−6 C)(0.25 m)2F = (8.987 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2) \frac{(4.3 \times 10^{-6} \text{ C})(4.3 \times 10^{-6} \text{ C})}{(0.25 \text{ m})^2} F=(8.987×109)1.849×10−110.0625 NF = (8.987 \times 10^9) \frac{1.849 \times 10^{-11}}{0.0625} \text{ N} F=(8.987×109)(2.9584×10−10) NF = (8.987 \times 10^9) (2.9584 \times 10^{-10}) \text{ N} F=2.66 NF = 2.66 \text{ N}

    The size of the force is 2.66 N\boxed{2.66 \text{ N}} and the type of force is repulsive\boxed{\text{repulsive}}.

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