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

Question

As shown in the figure above, 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 = 200 C. When the ball reaches point B, it travels horizontally there after. The height from A to B is h = 10 m and the horizontal distance between A and B is d = 1 m. You can ignore friction and use g = 10 m/s2 for your calculation.

A) How much is the kinetic energy of the ball when it reaches at B (in the unit of J)?

B) What is the minimum strength of electric field E that can stop the particle right at Point B?

E d HB As shown in the figure above a negatively charged ball is placed at point A and slides down the slope from rest. The area has a uniform electric field F 3 N/Cr pointing to the right. The mass of the ball ls m 10 kg and the charge is 200 C. When the ball reaches polnt B, It travels horizontally there after The helght from A to B ls h 10 m and the horizontal distance between A and BIs d 1 m. You can ignore friction and use g 10 m/s for your calculation. How much is the kinetic energy of the ball when it reaches at B (in the unit of J)? 1600 Submit Answer Incorrect. Tries 1/2 Previous Ilies what is the minimum strength of electric field L that can stop the particle right at Point B? Nyc

Explanation / Answer

Here, field is towards right so the ball experiences force towards left.

Therefore, force due to field = Eq (towards left)
=> mgsin(theta) - Eqcos(theta) = ma

Again, theta = arctan(10/1) = 84.3 deg

So, a = (10*10*0.99 - 3*200*0.10)/10 = 3.0 m/s^2

and s = sqroot(10^2 + 1^2) = 10.05

Further, use the formula -
2*a*s = v^2

(A) K.E. at point B = 0.5m*2*a*s = m*a*s = 10*3.0*10.05 = 301.5 J

(B) 2. K.E. = m*a*s = s*[mgsin(theta) - Eqcos(theta)] = 0
mgsin(theta) = Eqcos(theta)

=> E = (m*g*tan(theta)) / q = 10*10*tan(84.3)/200 = 100*10/200 = 5.0 N/C

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