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CAPA 4 Questions And Answers WellElaborated Rated A+
1
1. A metal sphere has a charge of +7.87mC (microC). What is the net charge
after 6.19×10¹³ electrons have been placed on it?
Answer: -2.03 μC
Explanation:
The charge of one electron is approximately −1.6×10−19-1.6 \times 10^{-
19}−1.6×10−19 C.
Total charge = 6.19×10136.19 \times 10^{13}6.19×1013 electrons
×−1.6×10−19\times -1.6 \times 10^{-19}×−1.6×10−19 C/electron
Total charge = −9.90×10−6-9.90 \times 10^{-6}−9.90×10−6 C.
The initial charge is +7.87 mC, so the net charge =
+7.87 mC−9.90 μC=−2.03 μC+7.87 \, \text{mC} - 9.90 \, \mu\text{C} = -2.03
\, \mu\text{C}+7.87mC−9.90μC=−2.03μC.
2. Calculate the difference in mass between two objects charged by induction.
Answer: 3.53 × 10⁻¹⁷ kg
Explanation:
The energy required to charge the objects comes from the mass-energy
equivalence.
The change in mass can be found using E=mc2E = mc^2E=mc2 for the
potential energy stored in the charges.
Mass difference = 3.53×10−173.53 \times 10^{-17}3.53×10−17 kg.
3. In a vacuum, two particles have charges of q₁ = +3.98mC and are separated by
0.278m. Particle 1 experiences an attractive force of 3.60N. What is the
magnitude of q₂?
Answer: 7.78 × 10⁻⁶ C
, Explanation:
Using Coulomb’s law:
F=k∣q1q2∣r2F = k \frac{|q₁ q₂|}{r^2}F=kr2∣q1q2∣
Solving for q2q₂q2, we get q2=7.78×10−6q₂ = 7.78 \times 10^{-6}q2
=7.78×10−6 C.
4. Two tiny conducting spheres carry charges of -21.1mC and +49.6mC,
separated by 2.60cm. What is the magnitude of the force on each sphere?
Answer: 1.39 × 10⁴ N
Explanation:
Using Coulomb’s law:
F=k∣q1q2∣r2F = k \frac{|q₁ q₂|}{r^2}F=kr2∣q1q2∣
Where r=2.60 cmr = 2.60 \, \text{cm}r=2.60cm, we get a force of
1.39×1041.39 \times 10^{4}1.39×104 N.
5. A charge of -3.01mC is fixed at the center of a compass. Two additional
charges are fixed on the circle of the compass (radius = 0.100m). The charges on
the circle are -3.97mC and +5.12mC. What is the magnitude of the net
electrostatic force acting on the charge at the center?
Answer: 1.75 × 10¹ N
Explanation:
Using Coulomb’s law, calculate the force between the central charge and
the two charges on the compass. The net force is the vector sum of the
individual forces.
6. A point charge of -0.672mC is fixed to one corner of a square, and identical
charges are fixed to the diagonally opposite corner. A point charge q is fixed to
each of the remaining corners. The net force on either charge q is zero. Calculate
the magnitude of q.
Answer: 1.90 × 10⁻⁶ C
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