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) $3q/{\epsilon }_0$

(2) $2q/{\epsilon }_0$

(3) $q/{\epsilon }_0$ 

(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

A solid metallic sphere has a charge +3Q. Concentric with this sphere is a conducting spherical shell having charge –Q. The radius of the sphere is a and that of the spherical shell is b (b > a). What is the electric field at a distance R(a < R < b) from the centre 

(1) $\frac{Q}{2\pi {\epsilon }_{0}R}$

(2) $\frac{3Q}{2\pi {\epsilon }_{0}R}$

(3) $\frac{3Q}{4\pi {\epsilon }_0{R}^2}$

(4) $\frac{4Q}{4\pi {\epsilon }_0{R}^2}$ 

A point charge \(q\) is placed at a distance \(\frac{a}{2}\) directly above the centre of a square of side \(a\). The electric flux through the square (i.e. one face) is:
1. \(\frac{q}{\varepsilon_0}\)
2. \(\frac{q}{\pi\varepsilon_0}\)
3. \(\frac{q}{4\varepsilon_0}\)
4. \(\frac{q}{6\varepsilon_0}\)

The charge on 500 cc of water due to protons will be:

1. 6.0 × 10²⁷ C
2. 2.67 × 10⁷ C
3. 6 × 10²³ C
4. 1.67 × 10²³ C