Question

Two stationary point charges +3.00 nC and +2.00 nC are separated by a distance of 50.0 cm. An electron is released from rest at a point midway between the two charges and moves along the line connecting the two charges. What is the speed of the electron when it is 10.0 cm from the +3.00-nC charge?

Answer 1

Explanation:

Point charge = $q_{1} = +3.00 \times 10^{-9} C$

    $q_{2} = +2.00 \times 10^{-9} C$

Separation by distance = s = 50.0 cm = 0.5 m   (as 1 m = 100 cm)

Electric potential energy of electron at 0.10 m = $-9 \times 10^{9} \times (1.6 \times 10^{-19}) \times 10^{-9}[\frac{3.00}{0.1} + \frac{2.00}{0.4}]$

Electric potential energy of electron at 0.10 m = $-5.04 \times 10^{-17}$ J

Change in P.E. will be calculated as follows.

                P.E = $2.16 \times 10^{-17} J$

                 K.E = $2.16 \times 10^{-17} J$

Therefore, we will calculate the speed as follows.

         Speed (v) = $\sqrt{\frac{2KE}{m}}$

                v = $\sqrt{[\frac{2 \times 2.16 \times 10^{-17}}{9.1 \times 10^{-31}}]}$

                  = $6.890 \times 10^{6}$ m/s

when electron is 10.0 cm from the +3.00 nC charge then its speed is $6.890 \times 10^{6}$ m/s.