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