Question

An electron is accelerated from rest through a potential difference $\Delta V\(. If the electron reaches a speed of \)7.26 \times 10^{6} \mathrm{m} / \mathrm{s},$ what is the potential difference? Be sure to include the correct sign. (Does the electron move through an increase or a decrease in potential?)

Short answer

Answer: The electron moved through a potential difference of approximately -1174 V, and it went through a decrease in potential.

Step-by-step solution

Write down the known values

We know the following values: - The final speed of the electron, \(v_f = 7.26 \times 10^{6} \mathrm{m} / \mathrm{s}\) - The initial speed of the electron, \(v_i = 0 \mathrm{m} / \mathrm{s}\) (since it starts from rest) - The mass of the electron, \(m_e = 9.11 \times 10^{-31} \mathrm{kg}\) - The elementary charge (charge of an electron), \(e = -1.6 \times 10^{-19} \mathrm{C}\)

Use the work-energy principle

Since the electron is accelerated by an electric field, we can use the work-energy principle, which states that the work done (\(W\)) on an object equals the change in kinetic energy: $$W = \Delta KE$$ Since the work is done by the electric field, we can relate work to the potential difference: $$W = q \Delta V$$ Where \(q\) is the charge of the electron and \(\Delta V\) is the potential difference.

Calculate the change in kinetic energy

The change in kinetic energy can be calculated using the formula: $$\Delta KE = \frac{1}{2}m(v_f^2 - v_i^2)$$ Plugging in the known values, we get: $$\Delta KE = \frac{1}{2}(9.11 \times 10^{-31} \mathrm{kg})(7.26 \times 10^{6} \mathrm{m} / \mathrm{s})^2$$ $$\Delta KE = 1.879 \times 10^{-17} \mathrm{J}$$

Calculate the potential difference

Now that we have the change in kinetic energy, we can calculate the potential difference using the equation: $$\Delta V = \frac{W}{q}$$ Replacing \(W\) with \(\Delta KE\), we have: $$\Delta V = \frac{1.879 \times 10^{-17} \mathrm{J}}{-1.6 \times 10^{-19} \mathrm{C}}$$ $$\Delta V \approx -1174 \mathrm{V}$$

Determine the sign of the potential difference

Since the potential difference is negative, it means that the electron went through a decrease in potential. This is because the electric field accelerated the negatively charged electron towards the higher potential side, causing a decrease in potential as it gained kinetic energy.