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A point charge q is placed on the top of a cone of semi vertex angle theta.Find the electric flux through the base of the cone.Plz solve it sir

Ashutosh Sharma
6 years ago

(a)q(1−cosθ)/2∈0
Explanation :
Consider a Gaussian surface , a sphere with its centre at the top and radius the slant length of the cone . The flux through the whole sphere isq/∈0.Therefore the flux through the base of the cone is
ϕe=(S/S0)x(q/∈0)
S0=area of whole sphere
S = area of sphere below the base of the cone .
Because all those field lines which pass through the base of the cone will pass through the cap of sphere
Let R= radius of Gaussion sphere
S0=area of whole sphere =4πR2
S = area of sphere below the base of the cone which can be found by integration .
dS=(2πr)Rdα
dS=2πR2sinαdα
S=∫(θ,0)dS=∫(θ,0)2πR2sinαdα
=2πR2(1−cosθ)
The desired fluxϕe=S/S0x(q/∈0)
=(2πR2(1−cosθ)/4πR2)(q/∈0)
ϕe=q/2∈0(1−cosθ)
Siddhant Sharma
19 Points
4 years ago
(a)q(1−cosθ)/2∈0Explanation :Consider a Gaussian surface , a sphere with its centre at the top and radius the slant length of the cone . The flux through the whole sphere isq/∈0.Therefore the flux through the base of the cone isϕe=(S/S0)x(q/∈0)S0=area of whole sphereS = area of sphere below the base of the cone .Because all those field lines which pass through the base of the cone will pass through the cap of sphereLet R= radius of Gaussion sphereS0=area of whole sphere =4πR2S = area of sphere below the base of the cone which can be found by integration .dS=(2πr)RdαdS=2πR2sinαdαS=∫(θ,0)dS=∫(θ,0)2πR2sinαdα=2πR2(1−cosθ)The desired fluxϕe=S/S0x(q/∈0)=(2πR2(1−cosθ)/4πR2)(q/∈0)ϕe=q/2∈0(1−cosθ)