Question

A positively charged particle starts at rest 25 cm from a second positively charged particle, which is held stationary throughout the experiment. The first particle is released and accelerates directly away from the second particle. When the first particle has moved 25 cm, it has reached a velocity of `10(sqrt2)ms^(-1)` . What is the maximum velocity `(in xx 10 ms^(-1))` that the first particle will reach?

Answer 1

`kq^(2)((1)/(r)-(1)/(2r))=(1)/(2)m(v^(2))`
or `(kq^(2))/(2r)=(1)/(2)mV^(2)`
or `(kq^(2))/(r)=(1)/(2)m(sqrt(2v))^(2)=(1)/(2)m(V_("max")^(2))`
or `V_("max")=Vsqrt(2)=(10sqrt(2))sqrt(2)=20ms^(-1)=2xx10ms^(-1)`