Question

A small particle has charge -4.90 μC and mass 2.10×10⁻⁴ kg . It moves from point A, where the electric potential is VA = 210 V , to point B, where the electric potential VB = 750 V is greater than the potential at point A. The electric force is the only force acting on the particle. The particle has a speed of 4.60 m/s at point A.

Answer 1

You didn't mention what to find in this question so I assume that we need to find :

What is its speed at point B ?

ans)

Charge = q = -4.90 *10^-6 C

Mass = m = 2.10×10−4kg

Initial velocity = u = 4.60 m/s

Final velocity = v

1/2)mv^2 - (1/2)mu^2 = q (Va - Vb)

multiply by 2 abd divide by m on both sides

v^2 - u^2 = 2q (Va - Vb) /m

v^2 = u^2 + 2q (Vb - Va) /m

v = sq rt [ u^2 + 2q (Va - Vb) /m ]

v = sq rt [ 4.60^2 - 2*4.90*10^-6 (210 -750) / 2.10×10−4 ]

v = sq rt [ 21.26 + 9.80*10^-6 (540 ) / 2.10×10−4 ]

v = sq rt [ 21.26 + 25.2 ]

v = 6.8161 m/s ------------------------>>>>>>>>>>>>>>>>answer)