Question

if e= qx/(a^2 +x^2)^3/2 represent electric field on the axis uniformaly charged ring of radius a then the value of e is maximum for what value of X on the axis from the centre of ring??
1) x= +/- a/√2
2) X= +/- a
3) X= +/- a/2
4) X= +/- a/2√2
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Answer 1

Answer:

The diagram is :-

The electric field due to the ring an it &

axis at a distance x is given by:-

E=(x2+R2)3/2kqx

To find maximum electric field, we will use the concept of maximum and minimum :-

dxdE=kq(x2+R2)(x2+R2)3/2−3/2(x2+R2)1/2.2x2

Now, dxdE=0⇒(x2+R2)3/2=23×2x2(x2+R2)1/2

⇒x2+R2=3x2

⇒2x2=R2

⇒x2=2R2

⇒x=±2R

So E