Asmall particle has charge -2.00 uc and mass 1.50×10-4 kg. it moves from point a, where the electric potential is v(a) = 200 v, to point b, where the electric potential v(b) = 920 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 4.30 m/s at point a.

1.)what is its speed at point b?
2.)and is it moving faster or slower at b than at a?

.

Explanation:

Since no external force is acting on the system.

Therefore, Total energy remains constant before and after.

So, Total energy of system= energy due to potential applied+kinetic energy

(Here v=velocity ,V=potential ,q=charge and m=mass).

Putting values .

We get,  .

At point B charged particle is moving faster as compared to point A.

Hence, it is the required solution.

higher potential

explanation:

because it is in a high speed so it must move into a of high potential

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j