Charge is uniformly distributed around a ring of radius R and the resulting electric field magnitude E is measured along the ring's central axis (perpendicular to the plane of the ring). At what distance from the ring's center is E maximum? (Use any variable or symbol stated above as necessary.)

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Netta00 Netta00 · Answer 1 · 2019-10-11T07:14:51+02:00

Answer:

[tex]x=\dfrac{r}{\sqrt2}[/tex]

Explanation:

Given that

Radius =r

Electric filed =E

Q=Charge on the ring

The electric filed at distance x given as

[tex]E=K\dfrac{Q}{(r^2+x^2)^{3/2}}[/tex]

For maximum condition

[tex]\dfrac{dE}{dx}=0[/tex]

[tex]E=K{Q}{(r^2+x^2)^{-3/2}}[/tex]

[tex]\dfrac{dE}{dx}=K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}[/tex]

For maximum condition

[tex]\dfrac{dE}{dx}=0[/tex]

[tex]K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}=0[/tex]

[tex]r^2+x^2-3x^2=0[/tex]

[tex]x=\dfrac{r}{\sqrt2}[/tex]

At [tex]x=\dfrac{r}{\sqrt2}[/tex] the electric field will be maximum.