Question

Use the worked example above to help you solve this problem. A proton is released from rest at x2.00 cm in a constant electric field with magnitude 1.50 x 10 N/C pointing in the positive x-direction. (a) Assuming an initial speed of zero, find the speed of a proton at x-0.0800 m with a potential energy of -2.40 x 10-17 j. (Assume the potential energy at the point of release is zero.) m/s (b) An electron is now fired in the same direction from the same position. Find the initial speed of the electron (at x -2.00 cm) given that its speed has fallen by half when it reaches x = 0.190 m, a change in potential energy of 5.05 x 10-13. m/s EXERCISE HINTS: GETTING STARTED I I'M STUCK! The electron in part (b) travels from x = 0.190 m (where it has half the initial speed you previously calculated) to x =-0.300 m within the constant electric field. If there's a change in electric potential energy of -1.18 x 1016 J as it goes from x -0.190 m to x--0.300 m, find the electron's speed at x =-0.300 m. (Note: Use the values from the Practice It section. Account for the fact that the electron may turn around during its travel.) m/s Need HelpRead It

Answer 1