A particle is uncharged and is thrown vertically upward from ground level with a speed of 26.3...

Question:

A particle is uncharged and is thrown vertically upward from ground level with a speed of 26.3 m/s. As a result, it attains a maximum height h. The particle is then given a positive charge +q and reaches the same maximum height h when thrown vertically upward with a speed of 32.5 m/s. The electric potential at the height h exceeds the electric potential at ground level. Finally, the particle is given a negative charge -q.

Ignoring air resistance, determine the speed with which the negatively charged particle must be thrown vertically upward, so that it attains exactly the maximum height h. In all three situations, be sure to include the effect of gravity.

Conservation of Energy:

The Law of Conservation of Energy states that the total energy of a system will be conserved or will remain constant when all the forces acting on the system are balance. This means that the net external force acting on the system is equal to zero.

Given:

{eq}v = 26.3 \frac {m}{s} {/eq} Speed of uncharged particle

{eq}v_+ = 32.5 \frac {m}{s} {/eq} Speed of positively charged particle

Using the Conservation of Energy principle for the uncharged particle, its kinetic energy is equal to its potential energy when it reach its maximum height,

{eq}\frac {1}{2} m v^2 = mgh {/eq} (1)

We can also apply the conservation of energy principle for the positively charged particle where its electric potential energy is equal to the difference of its kinetic energy and gravitational potential energy as,

{eq}qV = \frac {1}{2} m v_+^2 - mgh {/eq} (2)

Lastly, applying the conservation of energy principle for the negatively charged particle,

{eq}qV = \frac {1}{2} m v_-^2 + mgh {/eq}

Isolating the kinetic energy term on one side of the equation, since we are to look for the speed of the negatively charge particle as,

{eq}\frac {1}{2} m v_-^2 = mgh - qV {/eq}

Substituting equations (1) and (2) into this equation,

{eq}\frac {1}{2} m v_-^2 = \frac {1}{2} m v^2 - (\frac {1}{2} m v_-^2 - mgh) {/eq}

{eq}\frac {1}{2} m v_-^2 = \frac {1}{2} m v^2 - (\frac {1}{2} m v_-^2 - \frac {1}{2} m v^2) {/eq}

{eq}\frac {1}{2} m v_-^2 = m v^2 - \frac {1}{2} m v_-^2 {/eq} Notice that the mass m cancels out on both sides of the equation.

Solving for the speed of the negatively charged particle,

{eq}v_- = \sqrt {2 v^2 - v_+^2} = \sqrt {2 (26.3)^2 - (32.5)^2} = 18.08673547 \frac {m}{s} = \boxed {18.1 \frac {m}{s}} {/eq} 