Given below are two statements:

Statement I : An electric dipole is placed at the centre of a hollow sphere. The flux of electric field through the sphere is zero but the electric field is not zero anywhere in the sphere.

Statement II : If R is the radius of a solid metallic sphere and Q be the total charge on it. The electric field at any point on the spherical surface of radius r (

In the light of the above statements, choose the correct answer from the option given below:

Option 1 -

Statement I is false but Statement II is true

Option 2 -

Both Statement I and Statement II are true

Option 3 -

Statement I is true but Statement II is false

Option 4 -

Both Statement I and Statement II are false

0 2 Views | Posted a month ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    a month ago
    Correct Option - 1


    Detailed Solution:

    All the charge given to a conducting sphere resides on outer surface.

Similar Questions for you

S
Syed Aquib Ur Rahman

Gauss Law is only concerned with the total enclosed charge that finally tells us the total flux. The charges outside may change field patterns. They not affect the total flux. It's actually incorrect to assume the field due to the external charges should also affect the flux through the Gaussian surface. 

S
Syed Aquib Ur Rahman

Gauss Law does not directly give the electric field in all cases. It can only be used in calculations for symmetrical surfaces: spherical, cylindrical, or planar. 

S
Syed Aquib Ur Rahman

The integral form of Gauss Law is considered as an indirect form and only in theory. It will still create a mathematical problem. The Gaussian surface passing through a discrete charge means it lies on the surface. Half of the electric flux is outside and half in. Not on the boundary. And we know Gauss' Law holds true only when there are closed surfaces. 

R
Raj Pandey

? E ? d s ? = 0 ? net   = ? in   - ? out   = 0

? in   = ? out  

V
Vishal Baghel

Using gauss law in differential form

. E = ρ ε 0

. ( ε 0 E ) = ρ

. D = ρ

( x i ^ + y j ^ + z k ^ ) . D = ρ

ρ = e x s i n y + e x s i n y + 2 = 2 Q = 2 ( V o l u m e ) = 4 × 1 0 9 c = 4 n c

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