Question

The maximum electric field intensity on the axis
of a uniformly charged ring of charge 9 and
radius R is​

Answer 1

Answer:

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

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