Question 1

A sphere encloses an electric dipole with charge \pm 3 	imes 10^{-6} C. What is the total electric flux across the sphere?

Accepted Answer

zero. By Gauss's law flux depends on the net enclosed charge. A dipole encloses +3 	imes 10^{-6} - 3 	imes 10^{-6} = 0, so the total flux is zero.

Question 2

Two parallel infinite line charges with linear charge densities +λ C/m and −λ C/m are placed at a distance of 2R in free space. What is the electric field mid-way between the two line charges?

Accepted Answer

\frac{\lambda}{\pi\varepsilon_0 R} N/C. Each line produces \frac{\lambda}{2\pi\varepsilon_0 R} at the midpoint, and both point the same way (away from +λ, towards −λ). They add:

E = \frac{\lambda}{2\pi\varepsilon_0 R} + \frac{\lambda}{2\pi\varepsilon_0 R} = \frac{\lambda}{\pi\varepsilon_0 R}

Question 3

A hollow metal sphere of radius R is uniformly charged. The electric field due to the sphere at a distance r from the centre

Accepted Answer

zero as r increases for r < R, decreases as r increases for r > R. Inside a hollow charged sphere (r < R) there is no enclosed charge, so E = 0 throughout. Outside (r > R) it behaves like a point charge, E = \frac{Q}{4\pi\varepsilon_0 r^2}, which decreases as r increases.

Question 4

The electric field in a certain region is acting radially outward and is given by E = Ar. A charge contained in a sphere of radius "a" centred at the origin of the field will be given by

Accepted Answer

4\pi\varepsilon_0 Aa^3. At r = a, E = Aa = \frac{1}{4\pi\varepsilon_0}\frac{q}{a^2}

q = 4\pi\varepsilon_0 Aa^3

Question 5

What is the flux through a cube of side a if a point charge of q is at one of its corners?

Accepted Answer

\frac{q}{8\varepsilon_0}. A charge at a corner is shared by 8 cubes, so the charge effectively enclosed is \frac{q}{8}:

\phi = \frac{q}{8\varepsilon_0}

Question 6

A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will

Accepted Answer

remain the same. By Gauss's law flux depends only on the net enclosed charge, not on the radius of the Gaussian surface, so it remains the same.

Question 7

A square surface of side L metre in the plane of the paper is placed in a uniform electric field E (V/m) acting in the same plane at an angle θ with the horizontal side of the square. The electric flux linked to the surface, in units of V·m, is

Accepted Answer

0. The field lies in the plane of the surface, so it is parallel to the surface (perpendicular to the area vector). Hence \phi = \vec{E}\cdot\vec{A} = 0.

Question 8

A thin conducting ring of radius R is given a charge +Q. The electric field at the centre O of the ring due to the charge on the part AKB of the ring is E (directed along KO). The electric field at the centre due to the charge on the remaining part ACDB of the ring is

Accepted Answer

E along OK. The net field at the centre of a complete uniformly charged ring is zero. So the field of part ACDB must exactly cancel the field E (along KO) of part AKB — hence it is E directed along OK (opposite direction).

Question 9

A hollow cylinder has a charge q coulomb within it. If φ is the electric flux (in units of volt·metre) associated with the curved surface B, the flux linked with the plane surface A (in volt·metre) will be

Accepted Answer

\frac{1}{2}\left(\frac{q}{\varepsilon_0} - \phi\right). Total flux \phi_A + \phi_B + \phi_C = \frac{q}{\varepsilon_0}. By symmetry the two plane faces carry equal flux (\phi_A = \phi_C), so 2\phi_A + \phi = \frac{q}{\varepsilon_0}, giving \phi_A = \frac{1}{2}\left(\frac{q}{\varepsilon_0} - \phi\right)

Question 10

A square surface of side L metre is in the plane of the paper. A uniform electric field E (V/m), also in the plane of the paper, is limited only to the lower half of the square surface. The electric flux in SI units associated with the surface is

Accepted Answer

zero. The field lies in the plane of the surface, so it is parallel to the surface (perpendicular to the area vector), and \phi = |E||ds|\cos 90° = 0.

Question 11

A charge q is located at the centre of a cube. The electric flux through any face is

Accepted Answer

\frac{4\pi q}{6(4\pi\varepsilon_0)}. Total flux = \frac{q}{\varepsilon_0} = \frac{4\pi q}{4\pi\varepsilon_0}. Divided equally among 6 faces: \frac{4\pi q}{6(4\pi\varepsilon_0)}

Question 12

A charge q µC is placed at the centre of a cube of side 0.1 m, then the electric flux diverging from each face of the cube is

Accepted Answer

\frac{q 	imes 10^{-6}}{6\varepsilon_0}. Total flux = \frac{q 	imes 10^{-6}}{\varepsilon_0}. Through each of the 6 faces: \frac{q 	imes 10^{-6}}{6\varepsilon_0}

Question 13

A charge q is placed at the corner of a cube of side a. The electric flux through the cube is

Accepted Answer

\frac{q}{8\varepsilon_0}. A charge at a corner is shared by 8 cubes, so the flux through this cube is \frac{1}{8}\cdot\frac{q}{\varepsilon_0} = \frac{q}{8\varepsilon_0}

Question 14

A hollow insulated conducting sphere is given a positive charge of 10 µC. What will be the electric field at the centre of the sphere if its radius is 2 m?

Accepted Answer

Zero. Charge resides on the outer surface; inside the sphere the enclosed charge is zero, so by Gauss's law the electric field at the centre (and everywhere inside) is zero.

Question 15

A point charge +q is placed at the mid-point of a cube of side L. The electric flux emerging from the cube is

Accepted Answer

\frac{q}{\varepsilon_0}. By Gauss's law the total flux through the closed cube equals \frac{q_{enclosed}}{\varepsilon_0} = \frac{q}{\varepsilon_0}

Question 16

The electric field strength in air at NTP is 3 	imes 10^6 V/m. The maximum charge that can be given to a spherical conductor of radius 3 m is

Accepted Answer

3 	imes 10^{-3} C. Q_{max} = 4\pi\varepsilon_0 R^2 E_{max} = \frac{R^2 E_{max}}{9 	imes 10^9} = \frac{3 	imes 3 	imes 3 	imes 10^6}{9 	imes 10^9} = 3 	imes 10^{-3} C

Question 17

Two spherical conductors A and B of radii 1 mm and 2 mm are separated by a distance of 5 cm and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres A and B is

Accepted Answer

4 : 1. On connection both spheres reach the same potential. Using E = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}, the book evaluates the ratio of surface fields as \frac{E_A}{E_B} = \left(\frac{r_B}{r_A}\right)^2 = \left(\frac{2}{1}\right)^2 = 4 : 1.

Electric Flux and Gauss's Law — NEET Physics PYQ MCQs with Solutions

Questions with solutions