Answer :
. Two identical conducting spheres A and B carry charges Q and 2Q respectively. They are separated by a distance much larger than their diameters. A third identical conducting sphere C is uncharged. Sphere C is first touched to A, then to B, and finally removed. As a result, the electrostatic force between A and B, which was originally F, becomes:
Answer: 5F / 16
Solution: The initial force between the two charges separated by a distance d is F = 2kQ2 / d2 .
After touching, the charges become Q/2 and 5Q/4 and the force is 5kQ2 / 8d 2 = 5F / 16 .
2. A particle with charge 2 μC is placed at the origin. Another particle, with charge 4 μC, is placed 2 m from the origin on the x-axis, and a third particle, with charge 2 μC, is placed 2 m from the origin on the y-axis. The magnitude of the force on the particle at the origin is:
Answer: 2.0 × 10−2 N
Solution: The force due to the first particle is kq q / d 2 = 0.0180 N in the x direction and the
force from the second particle is kq q / d 2 = 0.0090 in the y direction. The magnitude of the net 13 13
force is then 0.0182 + 0.0092 = 0.020 N.
3. A particle with positive charge Q is on the y-axis a distance 2a from the origin, and a particle with positive charge q is on the x-axis a distance d from the origin. The value of d for which the x component of the force on the second particle is the greatest is:
12 12
Prof. Paul Avery Prof. Zongan Qiu Oct. 1, 2014
Answer: 2a
Solution: The x component of the force is kQqcosθ / 4a + d , where cosθ = d / 4a + d .
Thus we want to maximize the quantity kQqd / 4a + d . Setting the first derivative to 0
Answer: 5F / 16
Solution: The initial force between the two charges separated by a distance d is F = 2kQ2 / d2 .
After touching, the charges become Q/2 and 5Q/4 and the force is 5kQ2 / 8d 2 = 5F / 16 .
2. A particle with charge 2 μC is placed at the origin. Another particle, with charge 4 μC, is placed 2 m from the origin on the x-axis, and a third particle, with charge 2 μC, is placed 2 m from the origin on the y-axis. The magnitude of the force on the particle at the origin is:
Answer: 2.0 × 10−2 N
Solution: The force due to the first particle is kq q / d 2 = 0.0180 N in the x direction and the
force from the second particle is kq q / d 2 = 0.0090 in the y direction. The magnitude of the net 13 13
force is then 0.0182 + 0.0092 = 0.020 N.
3. A particle with positive charge Q is on the y-axis a distance 2a from the origin, and a particle with positive charge q is on the x-axis a distance d from the origin. The value of d for which the x component of the force on the second particle is the greatest is:
12 12
Prof. Paul Avery Prof. Zongan Qiu Oct. 1, 2014
Answer: 2a
Solution: The x component of the force is kQqcosθ / 4a + d , where cosθ = d / 4a + d .
Thus we want to maximize the quantity kQqd / 4a + d . Setting the first derivative to 0
The electric potential energy of the given charged particle is 8.55 x 10⁻⁷ J.
Electric potential energy
The electric potential energy of the given charged particle is calculated as follows;
E = Fd
where;
- F is the force between the charges
- d is the distance between the charges
[tex]E = \frac{kq_1q_2}{r^2} \times r\\\\E = \frac{kq_1q_2}{r}[/tex]
Substitute the given parameter and solve for electric energy
[tex]E = \frac{(9\times 10^9) \times (5 \times 10^{-9}) \times (9.5 \times 10^{-9})}{0.5} \\\\\E = 8.55 \times 10^{-7} \ J\\\\[/tex]
Thus, the electric potential energy of the given charged particle is 8.55 x 10⁻⁷ J.
Learn more about electric potential energy here: https://brainly.com/question/14306881