1.

An another point charge q, thrown towards the stationary point charge Q at speed v, it returns at a distance r from Q. If 2v speed is given instead of v then distance of closest approach will be ………

Answer»

r
2r
`(r )/(2)`
`(r )/(4)`

Solution :Distance of closest approach to `r_(0)=(2Ze^(2))/(4piin_(0)K_(0))`
`:.r_(0)prop(1)/(K_(0))` where `K_(0)` is initial KINETIC energy
Now `K_(0)=(1)/(2)mv_(0)^(2)`
`:.K_(0)prop(1)/(v_(0)^(2))`
`:.r_(0)prop(1)/(v_(0)^(2))`
`:.((r_(0))_(2))/((r_(0))_(1))=((v_(01))/(v_(02)))^(2)=(v^(2))/((2V)^(2))=(1)/(4)`
`:.(r_(0))_(2)=(r)/(2)""[:.(r_(0))_(1)=r]`


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