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In the assembly as shown below, the potential difference across the plates is 4 volts.A positive particle of charge +4e is projected from the negative plate with an initial kinetic energy of 4ev and the negative particle of charge (-2e) is projected from the positive plate.Both the particles reach point 'A' with zero kinetic energy.Find the initial kinetic energy of the negative particle in eV. |
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Answer» Solution :For the positive particle , applying energy CONSERVATION initially and at a POINT A `K.E._1+P.E._1=K.E._1+P.E._1` `implies 4eV+(+4E)(0V)=0+(+4e)(x"VOLT") " " (x="potential at point A")` `implies x=1` volt Now applying energy conservation for the negative particle at point 'A' and initially `implies K.E._1+(-2e)(4V)=0+(-2e)(1"volt")` `K.E._1-8eV=-2eV` `implies K.E._1=6eV` |
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