Question

There is an electric field of magnitude \(150 . \mathrm{N} / \mathrm{C},\) directed vertically downward, near the surface of the Earth. Find the acceleration (magnitude and direction) of an electron released near the Earth's surface.

Short answer

Answer: The acceleration of the electron is 2.63 x 10^{13} m/s^2, directed upward (opposite to the direction of the electric field).

Step-by-step solution

Write down the given values

The electric field E is given as 150 N/C (vertically downward). The charge of an electron (q) is -1.6 x 10^{-19} C.

Calculate the force on the electron due to the electric field

The force F experienced by a charged particle in an electric field E is given by F = qE. Here, we have q = -1.6 x 10^{-19} C and E = 150 N/C. Therefore, F = (-1.6 x 10^{-19} C)(150 N/C) = -2.4 x 10^{-17} N. The negative sign indicates that the force is in the upward direction (opposite to the direction of the electric field).

Determine the acceleration of the electron

The acceleration (a) of the electron can be found using the formula a = F/m, where m is the mass of the electron. The mass of an electron is approximately 9.11 x 10^{-31} kg. Therefore, a = (-2.4 x 10^{-17} N) / (9.11 x 10^{-31} kg) = 2.63 x 10^{13} m/s^2. The acceleration of the electron is 2.63 x 10^{13} m/s^2, directed upward (opposite to the direction of the electric field).