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

Answer 2

Speed (v final) of the electron which is at  10.0 cm away from charge 1 is,

$v=7.52\times10^{6}\rm m/s$

What is Coulombs law?

According to the Coulombs law, the force of attraction of repulsion between charged two bodies is directly proportional to the product of charges of them and inversely proportional to the square of distance between them.

It can be given as,

$F=\dfrac{kq_1q_2}{r^2}$

Here, (k) is the coulombs constant,  (q1 and q2) is the charges of two bodies and (r) is the distance between the two charges.

The charge 1 has the magnitude 3.00 nC. The charge 2 has the magnitude 1.65 nC.  

As the electron is in the midway, the distance between both charges and electron is,

$r=\dfrac{0.50}{2}\\r=0.25\rm m$

The potential energy of the electron from coulombs law can be given as,

$U_1=\dfrac{ke}{r}(q_1+q_2)$

Put the values as,

$U_1=\dfrac{9\times10^{9}1.6\times10^{-19}}{0.25}(3.00\times10^{-9}+1.65\times10^{-9})\rm J$

The distance of the electron is 10 cm from the charge 1. Thus the distance of the electron from charge 2 is 40 cm (50-10).

The potential energy of the electron for this case can be given as,

$U_2=\dfrac{kq_1e}{r_1}+\dfrac{kq_2e}{r_2}$

Put the values as,

$U_2=\dfrac{9\times10^{9}\times3.00\times10^{-9}\times1.6\times10^{-19}}{0.1}+\dfrac{9\times10^{9}\times1.67\times10^{-9}\times1.6\times10^{-19}}{40}$

The final velocity of the electron is given by the conservation of energy as,

$v=\sqrt{2\dfrac{U_2-U_1}{9.11\times10^{-31}}$

Put the values, in the above equation, we get,

$v=7.52\times10^{6}\rm m/s$

Thus the speed (v final) of the electron which is at  10.0 cm away from the charge 1 is,

$v=7.52\times10^{6}\rm m/s$

learn more about the Coulombs law here;

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