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Difficulty: MediumElectrostatics and Electric Charges

Two identical insulated conducting spheres, XX and YY, carry initial charges of +8.0×10−9 C+8.0 \times 10^{-9}\text{ C} and −2.0×10−9 C-2.0 \times 10^{-9}\text{ C}, respectively. They are brought into brief contact and then separated by a distance of 0.3 m0.3\text{ m} in a vacuum. Taking Coulomb's constant k=9.0×109 N⋅m2C−2k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}, what is the magnitude of the electrostatic force between the two spheres after separation?

  1. 9.0×10−7 N9.0 \times 10^{-7}\text{ N}Answer
  2. B
    1.6×10−6 N1.6 \times 10^{-6}\text{ N}
  3. C
    2.5×10−6 N2.5 \times 10^{-6}\text{ N}
  4. D
    3.6×10−6 N3.6 \times 10^{-6}\text{ N}

Answer

The magnitude of the electrostatic force between the spheres after separation is 9.0×10−7 N9.0 \times 10^{-7}\text{ N}.
When two identical conducting spheres touch, the total charge conserves and divides equally between them. The net charge is (+8.0×10−9 C)+(−2.0×10−9 C)=+6.0×10−9 C(+8.0 \times 10^{-9}\text{ C}) + (-2.0 \times 10^{-9}\text{ C}) = +6.0 \times 10^{-9}\text{ C}, yielding +3.0×10−9 C+3.0 \times 10^{-9}\text{ C} on each sphere. Substituting these equal charges and distance 0.3 m0.3\text{ m} into Coulomb's law gives F=9.0×109×(3.0×10−9)20.09=9.0×10−7 NF = \frac{9.0 \times 10^9 \times (3.0 \times 10^{-9})^2}{0.09} = 9.0 \times 10^{-7}\text{ N}.

Step-by-Step Solution

1
Calculate the total net charge after contact and the charge on each identical sphere.
Net charge Qnet=(+8.0×10−9 C)+(−2.0×10−9 C)=+6.0×10−9 CQ_{net} = (+8.0 \times 10^{-9}\text{ C}) + (-2.0 \times 10^{-9}\text{ C}) = +6.0 \times 10^{-9}\text{ C}. Since the spheres are identical, the charge on each sphere is q=+6.0×10−9 C2=+3.0×10−9 Cq = \frac{+6.0 \times 10^{-9}\text{ C}}{2} = +3.0 \times 10^{-9}\text{ C}.
When identical conductors touch, total charge is conserved and redistributes equally between them.
2
Apply Coulomb's Law using the redistributed charge and separation distance.
F=kq1q2r2=(9.0×109)(3.0×10−9)(3.0×10−9)(0.3)2=8.1×10−70.09=9.0×10−7 NF = k \frac{q_1 q_2}{r^2} = (9.0 \times 10^9) \frac{(3.0 \times 10^{-9})(3.0 \times 10^{-9})}{(0.3)^2} = \frac{8.1 \times 10^{-7}}{0.09} = 9.0 \times 10^{-7}\text{ N}.
Coulomb's Law calculates the magnitude of electrostatic force between two point-like charges at a given distance.

Key Concept

Charge conservation, redistribution between identical conductors, and Coulomb's Law
Estimated Time:1m 30s
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Topics

Electrostatics and Electric Charges

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