InterviewSolution
Saved Bookmarks
| 1. |
As shown in figure a dust particle with mass m = 5.0 × 10–9 kg and charge q0 = 2.0 nC starts from rest at point a and moves in a straight line to point b . What is its speed v at point b? A. `26 ms^-1`B. `34 ms^-1`C. `46 ms^-1`D. `14 ms^-1` |
|
Answer» Correct Answer - c Apply conservation of mechanical energy between points `a and b` `(KE + PE)_a = (KE + PE)_b` `0 + (k(3 xx 10^-9)q_(0))/(0.01) - (k(3 xx 10^-9)q_0)/(0.02)` =`(1)/(2) mv^2 + (k( 3 xx 10^-9)q_0)/(0.02) - (k(3 xx 10^-9)q_0)/(0.01)` Put the values and get `v = 12 sqrt(15) = 46 ms^-1`. |
|