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. ` 2. 65 cms^ ( - 1 ) `
B. `3.65 cms^( - 1 ) `
C. ` 4.65 cms ^( -1 ) `
D. `5.65 cms^ ( -1 ) `

Answer

C

Explanation

According to conservation of energy, we get `K_(a)+U_(a)=K_(b)+U_(b)`
Here, `K_(a)=0` and the potential energies are `U_(a)=q'V_(a) and U_(b)=q'V_(b)`
`therefore 0+q'V_(a)=(1)/(2)mv^(2)+q'V_(b)`
`or v=sqrt((2q'(V_(a)-V_(b)))/( m))` `or v=sqrt((2q'(V_(a)-V_(b)))/( m))`
`V_(a)=(9.0xx10^Nm^(2)C^(-2)) ((3xx10^(-9)C)/(0.01m)+(-3xx10^(-9)C)/(0.02m))=1350V`
`F_(b)=(9.0xx10^(9)Nm^(2)C^(-2)) ((3xx10^(-9)C)/(0.02m)+(-3xx10^(-9)C)/(0.01m))=-1350V`
`therefore v=sqrt((2(2xx10^(-9)C)(2700V))/(5xx10^(-3)kg))=4.65xx10^(-2)ms^(-1)`