An electron with an initial speed of 5.00××1055 m/s enters a region in which there is an electric field directed along its direction of motion. Note: you'll need to look up the mass and charge of an electron. 1) If the electron travels 9.00 cm in the field before being stopped, what are the magnitude and direction of the electric field

Since the electron is moving in the direction of the electric field, then the electric force on it will act in the direction opposite to its displacement whose magnitude is given by:

[tex]F = eE[/tex]

Since the displacement of the electron opposite to the direction of electron, the work done by electric force will be given by:

[tex]W = - Fx\\\Rightarrow W = -eEx[/tex]

Since the electron enters the region of the uniform electric field with an initial speed u and travels a distance x to come to rest. The change in kinetic energy of the electron can be given by:

[tex]\Delta K = \dfrac{1}{2}m(v^2-u^2)\\\Delta K = \dfrac{1}{2}m((0)^2-u^2)\\\Delta K = -\dfrac{1}{2}mu^2\\[/tex]

Since only the electric force is doing work on the electron, then the change in kinetic energy will be equal to the work done by electric force. This is due to the work-energy theorem.

[tex]\therefore \Delta K = W\\\Rightarrow -\dfrac{1}{2}mu^2=-eEx\\\Rightarrow \dfrac{1}{2}mu^2=eEx\\\Rightarrow E=\dfrac{mu^2}{2ex}\\\Rightarrow E=\dfrac{9.1\times 10^{-31}\times (5\times 10^{5})^2}{2\times 1.6\times 10^{-19}\times 0.09}\\\Rightarrow E=7.90\ N/C[/tex]

Hence, the magnitude of electric field is given 7.90 N/C which point in the direction of the displacement of the electron.