![2.4 [5] As shown in the figure, a charged particle remains stationary between the two charged plates. The plate separation is 2.0 cm, the mass and charge of the particle is m - 4.0 x 101 kg and q 2.4 x 108 C. Find the potential difference between the plates. Eq mg 2.5 [10] Page 560, Question #9 2.6 [10] Page 560 Question #10 -2.7 [ 101 Page 562, Ouestion #46 (Bonus)](http://img.homeworklib.com/questions/8e74b690-6ff1-11ea-b287-2b21b160949a.png?x-oss-process=image/resize,w_560)
Homework Answers
Mass, m= 4* 10^(-13) kg
Charge of the particle, q= 2.4 *10^(-18) C
If the charged particle is stationary, qE = mg
E= (mg/ q) = ( 4*10^(-13) * 9.81 / (2.4 *10^(-18))) = (16.35 * 10^(5)) V/m
Separation between the plates, d= 2 cm = 0.02m
If the potential difference between the plates is V
E = (V/d)
V = E*d = ( 16.35*10^(5)*0.02) =32.7 *10^(3) V = 32.7 kV (ans)
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