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
+1 vote
104 views
in Physics by (93.7k points)
closed by
A positively charge sphere of radius `r_0` carries a volume charge density `rho`. A spherical cavity of radius `r_0//2` is then scooped out and left empty. `C_1` is the center of the sphere and `C_2` that of the cavity. What is the direction and magnitude of the electric field at point B?
image
A. `(17 rhor_(0))/(54in_(0))l eft`
B. `(rhor_(0))/(6in_(0))l eft`
C. `(17 rhor_(0))/(54in_(0))right`
D. `(rho r_(0))/(6in_(0))` right

1 Answer

0 votes
by (94.0k points)
selected by
 
Best answer
Correct Answer - a
Electric field on surface of a uniformly charged sphere is given by `(Q)/(4piepsilon_(0)R^(2))=(rhoR)/(3epsilon_(0))`
Electric field at outside point is given by
`E=(Q)/(4piepsilonr^(2))=(rhoR^(3))/(3 epsilon_(0)r^(2))`
`E_(B)= E_("whole sphere")-E_("cavity")`
`E_(B)= (rhor_(0))/(3epsilon_(0))-(rho(r_(0)/(2))^(3))/(3epsilon_(0)((3r_(0))/(2))^(2))=(16rhor_(0))/(54epsilon_(0))`

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

...