According to Gauss’ Theorem, electric field of an infinitely long straight wire is proportional to 

(1) r

(2) $\frac{1}{r^2}$

(3) $\frac{1}{r^3}$

(4) $\frac{1}{r}$

Electric charge is uniformly distributed along a long straight wire of radius \(1\) mm. The charge per cm length of the wire is \(Q\) coulomb. Another cylindrical surface of radius \(50\) cm and length \(1\) m symmetrically encloses the wire as shown in the figure. The total electric flux passing through the cylindrical surface is:

The S.I. unit of electric flux is 

(1) Weber

(2) Newton per coulomb

(3) Volt × metre

(4) Joule per coulomb

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: 

(1) $({\phi }_1+{\phi }_2){\epsilon }_0$

(2) $({\phi }_2-{\phi }_1){\epsilon }_0$

(3) $({\phi }_1+{\phi }_2)/{\epsilon }_0$

(4) $({\phi }_2-{\phi }_1)/{\epsilon }_0$ 

Shown below is a distribution of charges. The flux of electric field due to these charges through the surface \(S\) is:

      
1. \(3q/ \varepsilon_0\)
2. \(2q/ \varepsilon_0\)
3. \(q/ \varepsilon_0\)
4. zero

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

(1) q₂ only

(2) Only the positive charges

(3) All the charges

(4) +q₁ and – q₁ only

The electric flux for Gaussian surface A that encloses the charged particles in free space is (given q₁ = –14 nC, q₂ = 78.85 nC, q₃ = – 56 nC) 

(1) 10³ Nm² C^–1

(2) 10³ CN^-1 m^–2

(3) 6.32 × 10³ Nm² C^–1

(4) 6.32 × 10³ CN^-1 m^–2

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 

(1) Directly proportional to r²

(2) Directly proportional to r³

(3) Inversely proportional to r

(4) Inversely proportional to r²

A positively charged ball hangs from a silk thread. We put a positive test charge q₀ at a point and measure F/q₀, then it can be predicted that the electric field strength E 

(1) > F/q₀

(2) = F/q₀

(3) < F/q₀

(4) Cannot be estimated