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two point charges + 1n C and -4 nc are 1m apart in air . Find the positions along the line joining the two charges at which resultant potential is zero. |
Answer» SOLUTION :  `V= (1)/(4piepsilon_(n) ) (q)/(r )` Position between the CHANGES `V_(1) + V_(2)= 0` `(1)/(4piepsilon_(0)) (1xx10^(-9))/(X) +(-(1)/(4piepsilon_(0))(4XX10^(-9))/(1-x))=0` `(1)/(4piepsilon_(0))(1xx10^(-9))/(x) = (1)/(4piepsilon_(0)) (4xx10^(-9))/(1-x)` `4X=1-x` `4x = x=1` `5x=1` `x= 0.2m` position beyond the charges  `(1)/(4piepsilon_(0)) (1xx10^(-9))/(x) = (1)/(4piepsi_(0)) ( 4xx10^(-9))/( 1+ x)` `4x=1+X` `3X=1` `X= 0.333 m` |
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