As shown in the figure, a negatively charged ball is placed at point A and slide

Question

As shown in the figure, a negatively charged ball is placed at point A and slides down the slope from rest. The area has a uniform electric field E = 3 N/C, pointing to the right. The mass of the ball is m = 10 kg and the charge is q = 10 C. When the ball reaches point B, it travels horizontally there after. The height from A to B is h = 30 m and the horizontal distance between A and B is d = 10 m. You can ignore friction and use g = 10 m/s2 for your calculation.

* How much is the kinetic energy of the ball when it reaches at B (in the unit of J)? The answer is 2700 J, but I have no idea how to solve this.

Explanation / Answer

force on the charged particle=q*E

as here q is negative, force direction is opposite to the direction of E field.

hence work done will be negative.

work done=-force*distance along the force

=-magnitude of charge*electric field magnitude*horizontal distance

=-10*3*10

=-300 J

using work -energy principle:

initial potential energy+work done against electric field

=final kinetic energy

=>mass*g*height-300=final kinetic energy

==>final kinetic energy=10*10*30-300=3000-300=2700 J

part2:

let minimum strength of electric field so that the particle stops at B=E N/C

then work done=-q*E*horizontal distance

=-10*E*10=-100*E

as particle stops at B final kinetic energy=0

then initial potential energy+work done=0

==>3000-100*E=0

==>E=3000/100=30 N/C

so minimum strength of electric field E is 30 N/C.

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