1) A charge of -5mc is release from point A to point B whenelectric field is kep

Question

1) A charge of -5mc is release from point A to point B whenelectric field is kept constant. At point B its kinetic energy is3.5 J. What is the potential difference between two points A &B?

2) A particle of mass 10^-5 kg and charge 3 micro coulombs, movesdown in a gravitational field from point A to point B a distance of3 m. The kinetic energy of the particle decreases by3.0 mJ duringthis motion. What is the potential difference between points A andB in Kv?

3) What is the speed of a proton in m/s that has been acceleratedfrom rest through a potential difference of 3.0 Kv?

Can anyone help with these problems?

Explanation / Answer

1. W=KE=3.5J Vq=3.5 V=3.5/-5*10-3 V=700J 2. PE=mgh PE=10-5*9.8*3J The net work donne is W=PE-KE W=2.94*10-4-3*10-3J W=Vq V=W/3*10-6 V V/106=    KV 3. Vq=1/2*mv2 3000*1.602*10-19=1/2*1.6721*10-27*v2 v2=5.748*1011 v=7.581*105m/s Hence we get by it. "Hope this helps!Best of luck for the rest of yourcoursework." v=7.581*105m/s Hence we get by it. "Hope this helps!Best of luck for the rest of yourcoursework."

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