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elixir [45]
3 years ago
13

Charge is uniformly distributed around a ring of radius R and the resulting electric field magnitude E is measured along the rin

g'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.)

Physics
1 answer:
Arte-miy333 [17]3 years ago
3 0

Answer:

x=\dfrac{r}{\sqrt2}

Explanation:

Given that

Radius =r

Electric filed =E

Q=Charge on the ring

The electric filed at distance x given as

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

For maximum condition

\dfrac{dE}{dx}=0

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

\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}}

For maximum condition

\dfrac{dE}{dx}=0

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

r^2+x^2-3x^2=0

x=\dfrac{r}{\sqrt2}

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

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