Question #329530

A small particle has charge – 5.00 µC and mass 2.00 X 10-4 kg. It moves from point A, where



the electric potential is VA = +2.00 V, to point B, where the electric potential is VB = +8.00 V. The



electric force is the only force acting on the particle. The particle has speed 5.00 m/s at point A.



What is its speed at point B? Is it moving faster or slower at B than at A? Explain.

1
Expert's answer
2022-04-18T17:31:48-0400

φq=m2(vB2vA2),\varphi q=\frac m2(v_B^2-v_A^2),

vB=2φqm+vA2=5.03 ms.v_B=\sqrt{\frac{2\varphi q}m+v_A^2}=5.03~\frac ms.


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