Question

what is the flux related with the cone if the point charge is placed just below the vertex

Answer 1

Answer:Method 1 : For point charge, the Gaussian surface should

be spherical. Consider a Gaussian sphere with its center at the

apex and radius the slant length of the cone. The flux through

the whole sphere is q/ε0. Therefore, the flux through the base

of the cone is ϕE=(A/A0)q/ε0. Here, A0 is the area of the

whole sphere (4πR2), and A is the area of the sphere below the

base of the cone. Consider a differential ring of radius r and

thickness dr, then

dA=(2πr)Rdα=(2πRsinα)Rdα(asr=Rsinα)

A=∫θ0(2πR2)sinαdα=2πR2(1−cosθ)

The desired flux is

ϕE=(AA0)1ε0=(2πR2)(1−cosθ)4πR2(qε0)

=(1−cosθ)q2ε0

Method 2: The total solid angle around a point in space is 4π

steradian. The solid angle subtended by the base of the cone

at the apex of the cone is (2π(1−cosθ)). As the flux associated

with solid angle 4π is q/ε0 , the flux associated with the solid

angle 2π=(1−cosθ) is

ϕ=qε02π(1−cosθ)4π=q(1−cosθ)2ε0.