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?

Options

A. `26 ms^-1`
B. `34 ms^-1`
C. `46 ms^-1`
D. `14 ms^-1`

Answer

c

Explanation

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`.