A cone of base radius R and height h is located in a uniform electric field $\overrightarrow{E}$ parallel to its base. The electric flux entering the cone is :
- $\frac{1}{2}$ E h R
- E h R
- 2 E h R
- 4 E h R
The Correct Option is B
Solution and Explanation
$= E\left(\frac{1}{2}h\times2R\right)$
$= Ehr$
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Concepts Used:
Gauss Law
Gauss law states that the total amount of electric flux passing through any closed surface is directly proportional to the enclosed electric charge.
Gauss Law:
According to the Gauss law, the total flux linked with a closed surface is 1/ε0 times the charge enclosed by the closed surface.
For example, a point charge q is placed inside a cube of edge ‘a’. Now as per Gauss law, the flux through each face of the cube is q/6ε0.
Gauss Law Formula:
As per the Gauss theorem, the total charge enclosed in a closed surface is proportional to the total flux enclosed by the surface. Therefore, if ϕ is total flux and ϵ0 is electric constant, the total electric charge Q enclosed by the surface is;
Q = ϕ ϵ0
The Gauss law formula is expressed by;
ϕ = Q/ϵ0
Where,
Q = total charge within the given surface,
ε0 = the electric constant.