Last Activity: 6 Years ago
let the total charge on the ring is Q then
electric field at the axis of a charged ring is given by
E=kQx/(R2 + x2)3/2
to find maximum electric field we can use the concept of maxima and minima....
dE/dx = d/dx of {kQx/(R2 + x2)3/2 }
=kQ{ 1 .(R2 +x2)3/2 - 3/2 (R2 +x2)1/2. x }/(R2 +x2)3 (by applying quotient rule)
now on putting dE/dx =0 ,we get
x2 -3x/2+R2 =0
x2 -3x/2 +9/16 -9/16 +R2 =0
(x-3/4)2 = 9/16-R2
x={(9-16R2)1/2 +3 }/4 units
therefore at a distance x from center of the ring electric field is maximum
Prepraring for the competition made easy just by live online class.
Full Live Access
Study Material
Live Doubts Solving
Daily Class Assignments
Last Activity: 3 Years ago