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
8.5k views
in Physics by (24.1k points)

Determine the force F of interaction between two hemispheres of radius R touching each other along the equator if one hemisphere is uniformly charged with a surface density σ1 and the other with a surface density σ2.

1 Answer

+1 vote
by (34.4k points)
selected by
 
Best answer

We choose two small arbitrary elements belonging to the surfaces of the first and second hemispheres and having the areas ΔS1 a nd ΔS2. Let the separation between the two elements be r12. The force of interaction between the two elements can be determined from Coulomb's law:

In order to determine the total force of interaction between the hemispheres, we must, proceeding from the superposition principle, sum up the forces ΔF12 for all the interacting pairs of elements so that the resultant force of interaction between the hemispheres is

where the coefficient k is determined only by the geometry of the charge distribution and by the choice of the system of units. If the hemispheres were charged with the same surface density σ, the corresponding force of interaction between the hemispheres would be

where the coefficient k is the same as in the previous formula

Let us determine the force F. For this purpose, we consider the "upper" hemisphere. Its small surface element of area ΔS carries a charge Δq = σ ΔS and experiences the action of the electric field whose strength E' is equal to half the electric field strength produced by the sphere having a radius R and uniformly charged with the surface density σ. (We must exclude the part produced by the charge Δq itself from the electric field strength.) The force acting on the charge Δq is

end is directed along the normal to the surface element. In order to find the force acting on the upper hemisphere, it should be noted that according to the expression for the force ΔF the hemisphere as if experiences the action of an effective pressure p = σ2/(2ε0). Hence the resultant force acting on the upper hemisphere is

(although not only the "lower" hemisphere, but all the elements of the "upper" hemisphere make a contribution to the expression ΔF for the force acting on the surface element ΔS, the forces of interaction between the elements of the upper hemisphere will be cancelled out in the general expression for the force of interaction between the hemispheres obtained above).

Since F = kσ2, we obtain the following expression for the force of interaction between the hemispheres for the case when they have different surface charge densities:

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

...