\"A particle is uncharged and is thrown vertically upward from ground level with

Question

"A particle is uncharged and is thrown vertically upward from ground level with a speed of 27.8 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 29.4 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."

I understand that this question is an application of the law of conservation of energy. I have attempted to solve it by setting (mgh)1 + (1/2)mv12 = (mgh)2 + (1/2)mv23 + EPEa = EPEb but have not had much luck. What am I missing?

Explanation / Answer

let for the charge q ,

the change in the electric potential energy is del E

Now, as for the + q the change is positive and it gains the electric potential energy ,

for the -q charge it will loose the same amount of energy ,

Now, let the velocity in the case of -q charge is v m/s

Now, as the change in both the cases is same

0.5 * m * 29.4^2 - 0.5 * m * 27.8^2 = 0.5 * m * 27.8^2 - 0.5 * m *v^2

29.4^2 - 27.8^2 = 27.8^2 - v^2

solving for v

v = 26.1 m/s

the speed of -q charged particle is 26.1 m/s

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