Question

A particle is uncharged and is thrown vertically upward from ground level with a speed of 23.1 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 27.3 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.
m/s

Two point charges, +3.06 µC and -6.20 µC, are separated by 1.10 m. What is the electric potential midway between them?
V

Answer 1

PROBLEM 1:

Let us assume at the ground level the potential energy due to gravitation and the electric potential are zero.

Now, the total energy of the uncharged particle at the ground level is

And the energy of the uncharged particle at the maximum height h is (the kinetic energy at this point becomes zero)

The conservation of energy gives us

So the maximum height is 27.225 m.

Now, when the particle is charged

The energy of the positively charged particle at the ground level is

And the energy of the positively charged particle at height h is

Putting the value of h

Conservation of energy gives us

So, the potential at height is 105.84m/q.

Now,

The energy of the negatively charged particle at the ground level is

And the energy of the negatively charged particle at the height h is

Putting the values of h and V in the above equation

The conservation of energy gives us

So, the initial velocity of the negatively charged particle must be 17.94 m/s to reach the same height of h.

__________________________________________________________________________________________

PROBLEM 2:

The electric potential midway is

So, the potential is -51381.8 V.