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

Question

A particle is uncharged and is thrown vertically upward from ground level with a speed of 26.1 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 31.0 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

Explanation / Answer

Some details should be given about the source of the electric field. In the absence of any info about this, I'll consider that the potential at any point in space is not changing by changing the sign of q (the source of the electric field remains the same during the experiments); U(0) and U(h) are constant.

Total energy conservation law is written as:

- first situation (no charge): mv02/2=mgh (h=hmax) (1)

- second situation (+q): mv022/2+qU(0)=qU(h)+mgh (2)

- third situation (-q): mv032/2-qU(0)=-qU(h)+mgh (3)

Work in (2) and (3) (use (1)):

mv022/2=q[U(h)-U(0)]+mv02/2 (4)

mv032/2=-q[U(h)-U(0)]+mv02/2 (5)

Work with (4) and (5) and get: v032=2v02-v022 ~ 20 m/s.

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