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Charge is uniformly distributed in a space. The net flux passing through the surface of an imaginary cube of side "a" in the space is “ ​x ​". The net flux passing through the surface of an imaginary sphere of radius "a" in the space will be

Charge is uniformly distributed in a space. The net flux passing through the surface of an imaginary cube of side "a" in the space is “​x​". The net flux passing through the surface of an imaginary sphere of radius "a" in the space will be
 

Grade:12

3 Answers

Gautam Rao
16 Points
5 years ago
should the answer be pi*x?
external flux of a surface is given by : E.ds.
since, the flux through the cube would be E*a2 = x
therefore for a sphere,  the flux would be E.*(pi)*a2
which is equal to x*pi.
 
note: since E.ds is a dot product of vectors...The angle between the field and the surface is 0.
Psv Dan
37 Points
4 years ago
answer is 2 pi * x / 3
imagine a cube of side a circumscribred by a sphere
apply formula x=E.a
both e and a are parallel
theta=0
flux is maximum
 
Avinash kumar
11 Points
4 years ago
E.S =X (given for cube)E. 6a^2 = X (S of cube = 6a^2)Then E= X/6a^2 Now X/6a^2×a^3 (E.f per unit volume)Now as given in the questions the E.f is uniformly distributed so,E.f through 4/3πr^3(which is volume of Sphere) is (4πa^3/3) ×(X/6a^2 ×a^3)=Which is equal to 4πX/18a^2Now Flux= E.f× Area(4πX/18a^2)×4πa^2(Area of sphere) =8π^2X/9 (Answer)From my point of view!😊

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