Two point charges of +2.0 μC and -6.0 μC are located on the x-axis at x = -1.0 cm and x 12) = +2.0 cm respectively. Where should
1 answer:
Answer:
x = 0.006 m
Explanation:
The potential at one point is given by
V = k ∑
remember that the potential is to scale, let's apply to our case
V = k (q₁ / x₁ + q₂ / x₂ + q₃ / x)
in this case they indicate that the potential is zero
0 = k (2 10⁻⁶ / (- 1 10⁻²) + (-6 10⁻⁶) / 2 10⁻² + 3 10⁻⁶ / x)
3 / x = + 2 / 10⁻² + 3 / 10⁻²
3 / x = 500
x = 3/500
x = 0.006 m
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The velocity considered is
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Generally from the law of energy conservation we have that
Here is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as
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And is the change in electric potential energy from initial position to a position of distance R from the nucleus , this is mathematically represented as
Here is zero because the electric potential energy at the initial stage is zero so
So
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Generally considering from initial position to a position of distance from the nucleus
Here represented the distance of the proton from the nucleus where the velocity is
Generally from the law of energy conservation we have that
Here is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as
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And is the change in electric potential energy from initial position to a position of distance
from the nucleus , this is mathematically represented as
Here is zero because the electric potential energy at the initial stage is zero so
So
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Divide equation 2 by equation 1
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