A charge +q is ejected from infinity with velocity v towards +Q charge ,it stops at a distance of R from the +Q charge.At what distance it will stop if it is ejected with 2v velocity

Dear Student,

In the Ist case, applying the conservation of the energy
Potential energy of change +q when it stop at distance R from change +Q = Initial kinetic energy of the charge +q
Qq4πε0R=12mqv2        ....(i)

Consider that, when change +q is projected with velocity 2v is stop at distance R' from charge +Q.
 applying the conservation of the energy, we get
Qq4πε0R'=12mq(2v)2        ....(ii)

Dividing equation (i) by (ii), we get


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Since kinetic energy during launch is converted to potential energy at a distance R
Therefore,When the charge is ejected with a velocity v,
                  1/2(mv2)=kQq/R ---------(1)
               When the charge is ejected with a velocity 2v,
                1/2(m(2v)2)=kQq/r --------(2)
     Divide eq. 1 and 2
    you will get r =R/4
Hope this helps
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