Question

1. An electron (q = -1.6x10-19 C, m=9.11x10-31 kg) is placed in an electric field of magnitude 3.52x104N/C which points in the positive z direction. Calculate the magnitude and direction of the acceleration of the electron.

2. An electron is initially moving in the +z direction, with a speed v=2.54x105 m/s at z=0. In the region between z=+1.50 cm and z=+2.50 cm, there is a uniform electric field of magnitude 4.10x104 N/C in the +x direction. Outside that region (z < +1.50 cm and > +2.50 cm) the electric field is zero. Find the velocity of the electron after it leaves the field region (z > +2.50 cm.)

3. In lab, you will use a device that accelerates electrons using electric fields generated between two conductive plates. A large electric field is used to "fire" the electrons out in a straight path. The electrons start from rest, accelerate a distance x (the distance between the conductive plates) through a field of strength Ex. Using the equation for the force on a charge in an electric field, Newton's 2nd Law of motion, and basic kinematics, find an equation for the final speed vx in terms of e, Ex , me, and x. (e is the charge of an electron and me is the mass of the electron.)

Answer 1

stutn 3/ ID 'kg 3.52 X107 N/ z G I82 XID 9-11 x)b 3) Dineution of acceleration along ngatire Valany Z 1.50 cm く2 く+2.500m Z 2.50um Zanis nauy 72-ad's FmA he x anis. M/s 7.200 I0 negion t = V Time taken Cho 1.00メ10 3.937 XD 2 54 x 1b vlori y ang me

ant 2.83510 Me = V Final velotityf cleitron 2 mls V 2. 35 x10 F E FxEA 3. 2 Ex, Me 2R S V; 2 Ex me