Problem 11: An electron with an initial speed of 29 × 10^{5} m/s is fired in the same direction as a uniform electric field of 80 N/C. How far does the electron travel before brought to rest and turned back?

Solution

Since electron has negative charge, therefore, it will be retarded in the given conditions. Using the equation F = q_{0}E (Force on the electron in the electric field) and then equating with F = ma. Remember that charge on an electron is 1.6 × 10^{-19 }C.

Now mass of an electron is 9.1 × 10^{-31} Kg, therefore,

Here the negative sign accounts for the direction of the acceleration (retardation).

Now to find the distance the electron covers before coming to rest (v_{f} = 0) can be calculated by using the equation of motion, 2as = v_{f}^{2} – v_{i}^{2}. Put the values,

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