Question

Two stationary positive point charges, charge 1 of magnitude 3.95 nC and charge 2 of magnitude 1.80 nC, are separated by a distance of 39.0 cm. An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.What is the speed of the electron when it is 10.0 cm from charge 1?

Answer 1

the speed of the electron from charge q1 is 7.17×10⁶ m/s

Step-by-step explanation:

Given the data in the question;

the potential at the center of the two charges will be;

V = k( q1/(d/2) + q2/(d/2)

so we substitute

V = (9×10⁹)( (3.95×10⁻⁹/(0.39/2) + 1.80×10⁻⁹/(0.39/2)

V = 265.4 V

the potential at a distance of 10 cm from the charges will be

V = k( q1/(d1) + q2/(d2)

(d1 = 10cm = 0.1m and d2 = 39cm - 10cm = 29cm = 0.29m )

V' = (9×10⁹)( (3.95×10⁻⁹/0.1 + 1.80×10⁻⁹/0.29

V' = 411.4 V

Now, from the conservation of energy the speed of the electron from charge q1 will be;

E = ( V' - V) qe

1/2mv² = ( V' - V) qe

v² = [( V' - V) qe] / 1/2m

v =√ ([( V' - V) qe] / 1/2m)

v =√ ([2( V' - V) qe] / m)

we substitute

v =√ (2[( 411.4 - 265.4) 1.6×10⁻¹⁹] / 9.1×10³¹)

v = 7.17×10⁶ m/s

Therefore, the speed of the electron from charge q1 is 7.17×10⁶ m/s