CBSE Class 12-science Answered
Two equal point charges A and N each of +1000esu are placed 200cm apart in air. A particle carrying a charge of -1000esu is projected along perpendicular bisector of AB from the point C midway between A &B , with a kinetic energy of 10^4erg. How far will the particle move before it turns back?
Asked by mridulabarua05 | 01 Feb, 2019, 12:57: PM
Expert Answer
First let us do units conversion so that all quantities are expressed in SI units
1000 esu of charge = 1000×3.336×10-10 C = 0.33 μC
10000 erg = 10000×10-7 J = 1 mJ
-0.33 μC of charge is projected with energy 1 mJ at a point midway between two charges 0.33 μC
which are placed 0.2 m apart, along the direction of perpendicular bisector of line connecting the charges.
When the projected charge moving along the perpendicular direction, its potential energy increases
till all its initial energy is converted to potential energy.
when it reaches the farthest point that can move, let r be the distance between the moving charge and each stationary charge.
By equating final potential energy to initial kinetic energy,
we have 
Hence farthest distance 
Answered by Thiyagarajan K | 01 Feb, 2019, 02:31: PM
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