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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
answered-by-expert 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 begin mathsize 12px style space 2 fraction numerator open parentheses 0.33 cross times 10 to the power of negative 6 end exponent close parentheses squared over denominator 4 straight pi cross times 8.854 cross times 10 to the power of negative 12 end exponent cross times straight r end fraction space equals space 10 to the power of negative 3 end exponent space J space space space space space o r space space space space r space equals space fraction numerator 2 cross times 0.33 cross times 0.33 over denominator 4 straight pi cross times 8.854 end fraction cross times 10 cubed space equals space 1.96 space m end style
Hence farthest distance begin mathsize 12px style square root of 1.96 cross times 1.96 space minus space 0.1 cross times 0.1 end root space almost equal to space 1.96 space m end style
Answered by Thiyagarajan K | 01 Feb, 2019, 02:31: PM

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