Question

A point charge q is kept at the center of circular plane surface of cone. A shaded section ABDCA shown at the slant surface of cone. If electric flux through the shaded section is
m 80x where n and m minimum possible integer. Find (n + m). [length AB=

Answer 1

A point charge q is kept at the centre of circular plane of cone. A shaded section ABCDA shown at the slant surface of cone. if the electric flux throught the shaded section is  $\frac{n}{m}\times\frac{q}{80\pi\epsilon_0}$ , where m and n are minimum possible integer.

we have to find the value of (m + n).

see the attached diagram,

from ΔpqR,

⇒ $\bf\frac{r}{h}=tan53^{\circ}$

⇒ $\bf r=\frac{4h}{3}$  [ ∵ tan53° = 4/3 ]

now first of all, we have to find the flux through sectional part ( see part 3).

now flux through the slant surface of cone containing the shaded section is given by , Ф = $\bf\frac{q}{2\epsilon_0}(1-cos53^{\circ})-\frac{q}{2\epsilon_0}(1-cos37^{\circ})$

$\bf\frac{q}{2\epsilon_0}\left[1-\frac{3}{5}\right]-\frac{q}{\epsilon_0}\left[1-\frac{4}{5}\right]\\\\=\bf\frac{q}{2\epsilon_0}\left[\frac{2}{5}-\frac{1}{5}\right]\\\\=\bf\frac{q}{10\epsilon_0}$

∵ electric flux through 2π(4h/3) = q/2ε₀

∴ electric flux through the shaded section , (AB = h/2) = $\frac{\frac{q}{10\epsilon_0}.\frac{h}{2}}{2\pi\frac{4h}{3}}$ = $\frac{3q}{160\pi\epsilon_0}$

after comparing to $\frac{n}{m}\times\frac{q}{80\pi\epsilon_0}$, we get m = 2 and n = 3

therefore the value of (m + n) = 5