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CBSE Class 12-science Answered

Two point charges of 1 microcoulomb and -0.25 microcoulomb are placed in air at a distance of 0.4m from each other. Find out at which point on the line joining the 2 charges should a third charge be placed so that no force acts upon it?
Asked by kandappan | 19 Mar, 2020, 04:34: PM
answered-by-expert Expert Answer
Let us consider where we put the test charge q. If we put the test charge in between 1μC and -0.25μC ,
direction of electrostatic forces on test charge q due to 1μC charge and -0.25μC charge will be same .
Hence forces will not be balanced.
 
If we put the test charge on left side of 1μC , force of repulsion will always greater than force of attraction.
Hence if we put the test charge on left side of 1μC, we can not balance the force.
 
If we put the test charge on right side of -0.25 μC , it is possible to balance the forces.
 
Let r be the distance from -0.25 μC , where the test charge is placed.
At this location sign of electrostatic forces on test charge q due to 1 μC and 0.25 μC are opposite.
 
Hence we need to find the distance r so that magnitude of these electrostatic forces are equal
 
begin mathsize 14px style fraction numerator q over denominator 4 pi epsilon subscript o left parenthesis 0.4 plus r right parenthesis squared end fraction space equals space fraction numerator 0.25 cross times q over denominator 4 pi epsilon subscript o space r squared end fraction end style
 
By solving above equation, we get r = 0.4 m
 
Answered by Thiyagarajan K | 19 Mar, 2020, 08:17: PM

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