A conic surface is placed in a uniform electric field E as shown in figure such that the field is perpendicular to the surface on the side AB. The base of the cone is of radius R, and the height of the cone is h. The angle of the cone is $θ$ . Find the magnitude of the flux that enters the cone's curved surface from the left side. Do not count the outgoing flux $(θ < 45^{\circ})$ .

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$ER [h \cos θ + πR \sin θ]$

$ER [h \sin θ + πR / 2 \cos θ]$

$ER [h \cos θ + π (R / 2) \sin θ]$

none of these

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Flux entering the cone from side AB will ultimately also pass through areas A₁ and A₂. So,

$ϕ = {EA}_{1} \cos θ + {EA}_{2} \cos (90 - θ)$

$= E (\frac{1}{2} 2 Rh \cos θ + \frac{π}{2} R^{2} \sin θ)$

$= ER [h \cos θ + π (R / 2) \sin θ]$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE