A point charge q is placed at the vertex of cone of semivertical angle 37- as shown in figure- Flux associated with half cone A B C O is n q- then n is Write upto two digits after the decimal point-

Θ = 37^o So the solid angle Ω is given by, nbsp;Ω = 2π (1 cos 37^o) = 2 π (1 4/5) = 2π/5 nbsp; ϕ _net = q/ϵ _o⇒ϕ = ϕ _net/4π . Ω = q/4πϵ _o . 2 π/5 = q ...

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Standard XII

Physics

Pressure

A point charg...

Question

A point charge q is placed at the vertex of cone of semivertical angle $37^{o}$ as shown in figure. Flux associated with half cone ABCO is $\frac{n q}{ϵ_{o}}$ then n is (Write upto two digits after the decimal point.)

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Solution

$θ = 37^{o}$
So the solid angle $Ω$ is given by,
$Ω = 2 π (1 - cos 37^{o}) = 2 π (1 - \frac{4}{5}) = \frac{2 π}{5}$
$ϕ_{n e t} = \frac{q}{ϵ_{o}} \Rightarrow ϕ = \frac{ϕ_{n e t}}{4 π} . Ω = \frac{q}{4 π ϵ_{o}} . \frac{2 π}{5} = \frac{q}{10 ϵ_{o}}$

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A point charge q is placed at the vertex of cone of semivertical angle 37o as shown in figure. Flux associated with half cone ABCO is nqϵo then n is (Write upto two digits after the decimal point.)

θ=37o So the solid angle Ω is given by, Ω=2π(1−cos 37o)=2π(1−45)=2π5 ϕnet=qϵo⇒ϕ=ϕnet4π.Ω=q4πϵo.2π5=q10ϵo

A point charge q is placed at the vertex of cone of semivertical angle $37^{o}$ as shown in figure. Flux associated with half cone ABCO is $\frac{n q}{ϵ_{o}}$ then n is (Write upto two digits after the decimal point.)

$θ = 37^{o}$
So the solid angle $Ω$ is given by,
$Ω = 2 π (1 - cos 37^{o}) = 2 π (1 - \frac{4}{5}) = \frac{2 π}{5}$
$ϕ_{n e t} = \frac{q}{ϵ_{o}} \Rightarrow ϕ = \frac{ϕ_{n e t}}{4 π} . Ω = \frac{q}{4 π ϵ_{o}} . \frac{2 π}{5} = \frac{q}{10 ϵ_{o}}$