Integrated Concepts An electron has an initial velocity of 5.00 × 10? m/s in a uniform 2.00 × 10? 5 N/C strength electric field. The field accelerates the electron in the direction opposite to its initial velocity. (a) What is the direction of the electric field? (b) How far does the electron travel before coming to rest? (c) How long does it take the electron to come to rest? (d) What is the electron’s velocity when it returns to its starting point?
Solution: 6 In the above question they have given the electron has an initial velocity of 5.00 × 10 5 m/s in a uniform 2.00 × 10 N/C strength electric field. The field accelerates the electron in the direction opposite to its initial velocity. Step 1 (a) What is the direction of the electric field The above figure diagram of the pattern of lines, sometimes referred to as electric field lines, point in the direction that a positive test charge would accelerate if placed upon the line. As the above such lines are directed away from positively charged source charges and toward negatively charged source charges. Step 2 (b) How far does the electron travel before coming to rest The coulomb’s law of electrostatic field, F = qE Where, q is charge of proton and E is the electric field Therefore Newton’s second law, (i.e) F = ma Where, m is the mass of proton and a is the acceleration of proton Step 3 In the above equation Newton’s second law and coulomb’s law of electrostatic field. We get, a = qE a = qE/m Then in the above question (b) we need to calculate the distance travelled electron before coming into rest So from Newton’s equation of motion, 2 2 v = v + 2axo Where, v is the final velocity of electron, v is the initial velocity of electron, o x is the distance travelled by electron, a is the acceleration of electron. We now need to substitute the 0 for v we get, 2 2 v = v + 2axo 2 0 = v o + 2ax In the above term the 0 is no value so we taking x in form of we get as, x = - v / 2a 2 o Then substitute q E/m for a we get as, x = - v m / oqE 5 In the above equation substitute 2.0 ×10 N/C for E , 9 10 -31 for q and 5 .0 ×10 m/s for v we get as, o x = (5.0 × 10 m/s) (9.11 ×10 = 35.58 × 10 m = 3.56 × 10 m Where the distance traveled by the electron before coming to rest is 3.56 × 10 m