Question

What is the speed vfinal of the electron when it is 10.0 cm from charge 1?

Answer 1

Answer:

Two stationary positive point charges, charge 1 of magnitude 3.45 nC and charge 2 of magnitude 1.85 nC, are separated by a distance of 50.0 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 v(final) of the electron when it is 10.0 cm from

The answer to the question is

The speed $v_{final}$ of the electron when it is 10.0 cm from charge Q₁

= 7.53×10⁶ m/s

Explanation:

To solve the question we have

Q₁ = 3.45 nC = 3.45 × 10⁻⁹C

Q₂ = 1.85 nC = 1.85 × 10⁻⁹ C

2·d = 50.0 cm

a = 10.0 cm

q = -1.6×10⁻¹⁹C

Also initial kinetic energy = 0 and

Initial electric potential energy = $k\frac{qQ_1}{d} + k\frac{qQ_2}{d} = kq(\frac{Q_1+Q_2}{d})$

Final kinetic energy due to motion = 0.5·m·v²

Final electric potential energy = $k\frac{qQ_1}{a} + k\frac{qQ_2}{2d-a} = kq(\frac{Q_1}{a}+\frac{ Q_2}{2d-a})$

From the energy conservation principle we have

$0+ kq(\frac{Q_1+Q_2}{d})=0.5mv^2+ kq(\frac{Q_1}{a}+\frac{ Q_2}{2d-a})$

Solving for v gives

$v=\sqrt{\frac{kq(\frac{Q_1+Q_2}{d})- kq(\frac{Q_1}{a}+\frac{ Q_2}{2d-a})}{0.5m}}$

where k = 9.0×10⁹ and m = 9.109×10⁻³¹ kg

gives v =7528188.32769 m/s or 7.53×10⁶ m/s

$v_{final}$ = 7.53×10⁶ m/s