Question

Point \(P\) is at a potential of \(500.0 \mathrm{kV}\) and point \(S\) is at a potential of \(200.0 \mathrm{kV} .\) The space between these points is evacuated. When a charge of \(+2 e\) moves from \(P\) to \(S\), by how much does its kinetic energy change?

Short answer

Answer: The change in kinetic energy of the charged particle is \(9.6\times10^{-14}\,\text{J}\).

Step-by-step solution

Understand the given parameters

We are given the following parameters: - Point P's potential: \(V_P = 500.0\,\text{kV}\) - Point S's potential: \(V_S = 200.0\,\text{kV}\) - Charge moving from P to S: \(q = +2e\) The electric potential difference between the two points can be found by subtracting the potential at Point S from the potential at Point P, which gives us \(\Delta V = V_P - V_S\).

Calculate the electric potential difference

We can now calculate the electric potential difference: \(\Delta V = V_P - V_S = (500.0 - 200.0)\,\text{kV} = 300.0\,\text{kV} = 3 \times 10^5\,\text{V}\)

Calculate the potential energy change

The change in the potential energy is given by the formula: \(\Delta U = q\Delta V\) Now, substitute the values of the charge (\(q = +2e = +2 \times 1.6\times10^{-19}\,\text{C}\)) and the electric potential difference (\(\Delta V = 3 \times 10^5\,\text{V}\)) to find the potential energy change: \(\Delta U = (+2\times1.6\times10^{-19}\,\text{C})(3\times10^5\,\text{V}) = 9.6\times10^{-14}\,\text{J}\) Since the charge is positive and moving from higher potential to a lower potential, it loses potential energy.

Apply the work-energy principle

The work-energy principle states that the change in kinetic energy is equal to the work done on a particle. In this case, the work done on the particle is equal to the change in potential energy: \(\Delta K = \Delta U\)

Find the change in kinetic energy

Using the result of the potential energy change: \(\Delta K = 9.6\times10^{-14}\,\text{J}\) The kinetic energy of the particle increases by \(9.6\times10^{-14}\,\text{J}\) when it moves from point P to point S.