Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
204 views
in Physics by (93.5k points)
closed by
A point charge Q is located on the axis of a disc of radius R at a distance b from the plane of the disc (figure). Show that if one-fourth of the electric flux from the charge passes through the disc, then `R=sqrt3b`.
image
A. `sqrt(5)b`
B. `sqrt(2)b`
C. `sqrt(3)b`
D. `2sqrt(3)b`

1 Answer

0 votes
by (93.9k points)
selected by
 
Best answer
Correct Answer - c
Total flux associated with the charge `Q` for solid angle `4p is (Q)/(epsilon_(0)`
Solid angle substended by disc at `Q`
`Omega= 2pi(1-cos alpha)= 2pi[1-(b)/(sqrt(b^(2)+R^(2)))]`
Flux through disc `= (Q Omega)/(epsilon_(0) 4pi)`
It is given to be `=(Q)/(4 epsilon_(0))`
So, `(Q)/(4epsilon_(0))=(Q Omega)/(epsilon_(0)4pi)`
Solve to get `R= sqrt(3)b`
image

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...