Physics - Flux and Gauss theorem

Electric field at a point varies as r⁰ for

• An electric dipole

• A point charge

• A plane infinite sheet of charge

• A line charge of infinite length

An electric charge is placed at the center of a cube of side 'a' . The electric flux on one of its faces will be

• $\frac{q}{6\in_0}$

• $\frac{q}{\in_0}$

• $\frac{q}{\in_0a^2}$

• zero

 Fig. shown is a distribution of charges. The flux of electric field due to these charges through the surface S is

• $\frac{3q}{\in_0}$

• $\frac{q}{\in_0}$

• zero

• $\frac{2q}{\in_0}$

The electric intensity due to an infinite cylinder of radius R and having charge q per unit length at a distance r (r>R) from its axis is

• Directly proportional to $r^2$

• Directly proportional to r

• Inversely proportional to r

• Inversely proportional to $r^2$

Consider the charge configuration and spherical Gaussian surface as shown in the figure. When calculating the flux of the electric field over the spherical surface the electric field will be due to

• All the charges

• Only the positive charges

• +q₁ and -q1

• q2

The S.I. unit of electric flux is

• Joule per coulomb

• Newton per coulomb

• Weber

• Volt metre

An electric dipole is put in north-south direction in a sphere filled with water. Which statement is correct

• Electric flux is coming towards sphere

• Electric flux is coming out of sphere

• Electric flux entering into sphere and leaving the sphere are same

• Water does not permit electric flux to enter into sphere

Eight dipoles of charges of magnitude are placed inside a cube. The total electric flux coming out of the cube will be

• $\frac{8e}{\in_0}$

• 0

• $\frac{16e}{\in_0}$

• $\frac{e}{\in_0}$

Two infinite plane parallel sheets separated by a distance 'd' have equal and opposite uniform charge densities  $\sigma$  . Electric field at a point between the sheets is 

• Zero

•  $\frac{\sigma}{2\in_0}$  

• Depends upon the location of the point

•  $\frac{\sigma}{\in_0}$  

If the electric flux entering and leaving an enclosed surface respectively is  $\phi_1$   and  $\phi_2$   the electric charge inside the surface will be 

•  $\left(\phi_1+\phi_2\right)\in_0$  

•  $\left(\phi_1-\phi_2\right)\in_0$  

•  $\frac{\left(\phi_1+\phi_2\right)}{\in_0}$  

•  $\frac{\left(\phi_1-\phi_2\right)}{\in_0}$