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negatively charged particle is released from rest at point B and accelerates until it reaches point A. The mass and charge of the particle are 4.0 106 kg and 2.0 105 C, respectively. Only the gravitational force and the electrostatic force act on the particle, which moves on a horizontal straight line without rotating. The electric potential at A is 36 V greater than that at B; in other words, VA VB 36 V. What is the translational speed of the particle at point A?

Respuesta :

Answer:

[tex]v = 18.97 m/s[/tex]

Explanation:

Since the charge particle is negative so it will move from lower potential to higher potential

So here we will have

[tex]-q(V_h - V_L) = \frac{1}{2}mv^2[/tex]

now we have

[tex]-q(V_B - V_A) = \frac{1}{2}mv^2[/tex]

[tex](-2 \times 10^{-5})(-36) = \frac{1}{2}(4\times 10^{-6})v^2[/tex]

[tex]7.2 \times 10^{-4} = 2\times 10^{-6} v^2[/tex]

[tex]v^2 = 360[/tex]

so we have

[tex]v = 18.97 m/s[/tex]

Answer:

Here some mistakes, the correct value of charge Q is -2.0x10⁻⁵ C and the mass is 4x10⁻⁶ kg.

The answer is 18.97 m/s

Explanation:

Please look at the solution in the attached Word file

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