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?

Explanation:

Point charge =

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

Electric potential energy of electron at 0.10 m =

Electric potential energy of electron at 0.10 m = J

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

                P.E =

                 K.E =

Therefore, we will calculate the speed as follows.

         Speed (v) =

                v =

                  = m/s

when electron is 10.0 cm from the +3.00 nC charge then its speed is m/s.

statement #3 on andre"s list is false.

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm