Question

22.6 a point charge in an electric field
an electron is accelerated from rest by a uniform electric field of magnitude 8.0 × 10^4 n/c. how long would it take for the electron to reach the speed of 8.0% of the speed of light?
o 1.7 ns
o 1.3 ns
o 27 ns
o 2.7 ns
o 13 ns

Explanation:

Step1: Determine the speed of the electron

The speed of light $c = 3\times10^{8}\ m/s$. The speed of the electron $v$ is $8.0\%$ of the speed of light, so $v=0.08\times3\times 10^{8}\ m/s = 2.4\times 10^{7}\ m/s$.

Step2: Calculate the force on the electron

The force on an electron in an electric - field is $F = qE$, where $q = 1.6\times10^{-19}\ C$ (charge of an electron) and $E = 8.0\times10^{4}\ N/C$. So $F=(1.6\times 10^{-19}\ C)\times(8.0\times 10^{4}\ N/C)=1.28\times 10^{-14}\ N$.

Step3: Calculate the acceleration of the electron

According to Newton's second law $F = ma$, and the mass of an electron $m = 9.1\times10^{-31}\ kg$. So the acceleration $a=\frac{F}{m}=\frac{1.28\times 10^{-14}\ N}{9.1\times 10^{-31}\ kg}\approx1.4\times 10^{16}\ m/s^{2}$.

Step4: Use the kinematic equation to find the time

The kinematic equation $v = v_0+at$, since the electron starts from rest ($v_0 = 0$), then $t=\frac{v - v_0}{a}=\frac{2.4\times 10^{7}\ m/s-0}{1.4\times 10^{16}\ m/s^{2}}\approx1.7\times 10^{-9}\ s = 1.7\ ns$.

Answer:

1.7 ns