As shown in fig. a dust particle with mass `m=5.0xx10^(-9)` kg and charge `q_(0)=2.0nC` 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. `26ms^(-1)`
B. `34ms^(-1)`
C. `46ms^(-1)`
D. `14ms^(-1)`

Answer

C

Explanation

Apply conservation of mechanical energy between points a and b
`(KE+PE)_(a)=(KE+PE)_(b)`
`0+(k(3xx10^(-9))q_(0))/(0.01)-(k(3xx10^(-9))q_(0))/0.02`
`=1/2mv^(2)+(k(3xx10^(-9))q_(0))/0.02-(k(3xx10^(-9))q_(0))/0.01`
Put the values and get `v=12sqrt(15)=46ms^(-1)`