A cone of base radius R and height h is located in a uniform electric field E parallel to its base- The electric flux entering the cone is-

The correct option is D E h R We #160;know #160;that, #160;ϕ #160;= #160;E.A #160;Cos #160;θ Flux #160;crossing #160;through #160;th ...

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Standard XII

Physics

Introducing Flux

A cone of bas...

Question

A cone of base radius $R$ and height $h$ is located in a uniform electric field $\to E$ parallel to its base. The electric flux entering the cone is:

4 E h R

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\frac{1}{2}

E h R

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2 E h R

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E h R

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Solution

The correct option is D $E h R$
We know that,
$ϕ = E . A C o s θ$
Flux crossing through the area of cone parallel to the electric field which is nothing but a right angle triangle having base $2 R$ and height $h$
therefore, $ϕ = E h R$

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A conic surface is placed in a uniform electric field $E$ as shown such that field is perpendicular to the surface on the side AB. The base of the cone is of radius $R$ and height of the cone is $h$ . The angle of cone is $θ$ as shown. Find the magnitude of that flux which enters the cone's curved surface on the left side. Don’t count the outgoing flux.

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A cone of base radius R and height h is located in a uniform electric field →E parallel to its base. The electric flux entering the cone is:

The correct option is D EhRWe know that,ϕ=E.ACosθFlux crossing through the area of cone parallel to the electric field which is nothing but a right angle triangle having base 2R and height htherefore, ϕ=EhR

A cone of base radius $R$ and height $h$ is located in a uniform electric field $\to E$ parallel to its base. The electric flux entering the cone is:

The correct option is D $E h R$
We know that,
$ϕ = E . A C o s θ$
Flux crossing through the area of cone parallel to the electric field which is nothing but a right angle triangle having base $2 R$ and height $h$
therefore, $ϕ = E h R$