Respuesta :
Answer:
Explanation:
Distance of second charge from origin d
d = [tex]\sqrt{(1.25)^ 2+(0.57)^ 2}[/tex]
= 1.374 m
electric potential energy of system of charge
= 9 x 10⁹ x 3.2 x 3.2 x 10⁻¹² / 1.374
67.07 x 10⁻³ J
This energy will be converted into kinetic energy
1/2 x m v² = 67 x 10⁻³
v² = 2 x 67 x 10⁻³ / m
= 2 x 67 x 10⁻³/ 2.63 x 10⁻³
v = 7.13 m /s
b ) kinetic energy at this point = 67/4 x 10⁻³
= 16.75 x 10⁻³
Potential energy = 67x10⁻³ - 16.75 x 10⁻³ = 50.75 x 10⁻³
If r be the required distance
9 x 10⁹ x 3.2 x 3.2 x 10⁻¹² / r = 50.75 x 10⁻³
r = 1.87 m
A second charge of 3.20 μC is released from rest at the position and first is at origin. PE is equal to KE for the second charge as it start to move from rest.
- a) The speed of second charge, when it moves infinitely far from the origin is 7.13 m/s.
- b) The distance from the origin does the second charge attain half the speed it will have at infinity is 1.87 meters.
What is electric force?
Electric force is the force of attraction of repulsion between two bodies.
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.
Given information-
The charge on first point which is held fixed at the origin is 3.20 μC.
The charge on second point which is at (1.25 m, 0.570 m) is 3.20 μC.
The mass of the second charge is 2.63 g ,
- a) The speed of second charge, when it moves infinitely far from the origin-
The shortest distance of the point (1.25 m, 0.570 m) from the first point (origin 0, 0) is,
[tex]d=\sqrt{1.25^2+0.57^2} =1.347[\tex]
The electric potential energy is equal to the kinetic energy for the second charge as it start to move from rest. Thus,
[tex]k\dfrac{q_1q_2}{r}=\dfrac{1}{2}mv^2[\tex]
Put the values as,
[tex]9\times10^9\dfrac{3.2\times10^{-6}\times3.2\times10^{-6}}{1.374}=\dfrac{1}{2}2.63\times10^{-3}v^2\\v=7.13[\tex]
Thus the speed of second charge, when it moves infinitely far from the origin is 7.13 m/s.
- b) The distance from the origin does the second charge attain half the speed it will have at infinity-
Potential energy is,
[tex]3\dfrac{9\times10^9\dfrac{3.2\times10^{-6}\times3.2\times10^{-6}}{1.374}}{4}=50.75\times10^{-3}[\tex]
Let the distance is r. Thus,
[tex]9\times10^9\dfrac{3.2\times10^{-6}\times3.2\times10^{-6}}{r}=50.75\times10^{-3}[\tex]
On solving the above equation we get,
r=1.87 m.
Thus the distance from the origin does the second charge attain half the speed it will have at infinity is 1.87 meters.
Hence,
- a) The speed of second charge, when it moves infinitely far from the origin is 7.13 m/s.
- b) The distance from the origin does the second charge attain half the speed it will have at infinity is 1.87 meters.
learn more about the electric force here;
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