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Home / Doubts and Solutions/JEE/Physics

Asked by raos9979 20th April 2021, 10:10 AM
Answered by Expert
Answer:
Let Length of wire be 2L and charge on wire is Q. Electric field at point P which is at the perpendicular bisector of wire
and at adistance y from mid point is given as
 
begin mathsize 14px style E space equals space K space cross times fraction numerator Q over denominator y space square root of y squared plus L squared end root end fraction end style
 
Where K = 1/ (4πεo ) = 9 × 109 N m2 C-2 is Coulomb's constant
 
Let us substitute values in above expression for electric field
 
Q = 5 × 10-9 C   ,   L = 1.5 m  , y = 2 m
 
begin mathsize 14px style E space equals space fraction numerator 9 cross times 10 to the power of 9 cross times 5 cross times 10 to the power of negative 9 end exponent over denominator 2 square root of 1.5 cross times 1.5 space plus space 2 cross times 2 end root end fraction space equals space 9 space V divided by m end style
Answered by Expert 20th April 2021, 2:25 PM
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