Near the surface of the Earth there is an electric field of about 150 {\rm V}/{\rm m} which points downward. Two identical balls with mass m = 0.490 kg are dropped from a height of 1.80 m, but one of the balls is positively charged with q_1 = 950 \mu C, and the second is negatively charged with q_2 = -95
0 \mu C.Use conservation of energy to determine the difference in the speed of the two balls when they hit the ground. (Neglect air resistance.)

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cjmejiab cjmejiab · Answer 1 · 2019-11-02T18:55:51+01:00

Answer:

0.175 m/s

Explanation:

We solve the problem for both negative and positive charges,

A due to Electric Field = [tex]\frac{qE}{m}[/tex]

Thus the speed of the negative charge is:

[tex]a1 = 9.8 - 950 x 10-6 x 150 / 0.49 = 9.51 m/s[/tex]

Thus the speed of the positive charge is:

[tex]a1 = 9.8 + 950 x 10-6 x 150 / 0.49 = 10.09 m/s[/tex]

Using the velocity difference equation then we have

[tex]mgh = \frac{1}{2}m*v^2[/tex]

[tex]v_1 - v_2 = \sqrt{((2)(10.09)(1.8)}- \sqrt{((2)(9.51)(1.8))} = 0.175 m/s[/tex]