Help in this question.

Asked by carnivalgirl8421 | 29th Jun, 2022, 10:06: PM

Expert Answer:

Total Flux Φ of electric field passing through the surface of cone is given as
 
begin mathsize 14px style ϕ space equals space surface integral E with rightwards arrow on top times stack d a with rightwards arrow on top end style ........................... (1)
where begin mathsize 14px style E with rightwards arrow on top end style is intensity of electric field at surface of cone and begin mathsize 14px style stack d a with rightwards arrow on top end style is small area element on the surface of cone.
Above equation implies that intensity of electric frield is integrated over the closed surface of cone.
 
By Gauss theorem, we have
 
begin mathsize 14px style ϕ space equals space surface integral E with rightwards arrow on top times stack d a with rightwards arrow on top space equals space q over epsilon subscript o end style  ........................... (2)
where q is enclosed charge inside the volume of cone.
 
In this case, enclosed charge is zero , because we are given that electric field intensity is 20begin mathsize 14px style i with hat on top end style N/C .
Given electric field intensity is uniform and directed along x-axis.
 
We will not get such a uniform intensity of electric field over the surface of cone
if there is an enclosed charge within the volume of cone.
 
Hence from eqn.(2) and by knowing that enclosed charge is zero ,
it can be concluded that total flux of electric field through the surface of cone is zero.
 
Physically we can infer that whatever flux entering from left side surface of cone is leaving
through right side surface of cone so that total flux of electric field over the surface of cone is zero.
 

Answered by Thiyagarajan K | 30th Jun, 2022, 08:08: AM