Question

A point charge 4.20μC is held fixed at the origin. A second point charge 1.40μC with mass of 2.80×10 −4kg is placed on the x axis, 0.280 m What is the electric potential energy U of the pair of charges? (Take U to be zero when the charges have infinite separation.) from the origin. Express your answer with the appropriate units. Part B The second point charge is released from rest. What is its speed when its distance from the origin is 0.500 m. Express your answer with the appropriate units. A point charge 4.20μC is held fixed at the origin. A Part C second point charge 1.40μC with mass of 2.80×10 −4kg is placed on the x axis, 0.280 m from the origin. What is its speed when its distance from the origin is 5.00 m. Express your answer with the appropriate units. Part D What is its speed when its distance from the origin is 50.0 m. Express your answer with the appropriate units.

Answer 1

Part A: The electric potential energy U of the pair of charges is
`\(-3.92 \, \text{J}\)`. Part B: The speed of the second point charge when its distance from the origin is
`\(0.500 \, \text{m}\) is \(1.65 \, \text{m/s}\)`. Part C: The speed of the second point charge when its distance from the origin is
`\(5.00 \, \text{m}\)` is
`\(0.33 \, \text{m/s}\)`. Part D: The speed of the second point charge when its distance from the origin is
`50.0 \, \text{m}\)` is
`\(0.033 \, \text{m/s}\)`.

Step-by-step explanation:

For Part A, the electric potential energy U of the charges is given by
`\(U = (k \cdot q_1 \cdot q_2)/(r)\)`, where k is Coulomb's constant,
`\(q_1\) and \(q_2\)` are the charges, and r is the separation distance. Plugging in the values, we get
`\(U = -3.92 \, \text{J}\).`

For Part B, the initial potential energy is converted into kinetic energy KE as the charge moves. Using the conservation of energy principle, KE = U, we can find the speed v using
`\(KE = (1)/(2)mv^2\).`

For Parts C and D, we can use the conservation of energy principle again to find the final speeds at different distances. The potential energy at the initial distance is converted into kinetic energy at the final distance.