CUET UG Physics Booster Test – 2 Flux and Gauss’s Law

Electric flux is: Given: Substituting: The angle used in the formula is always the angle between the electric field and the area vector. Since the area vector makes 60° with the field, the cosine factor reduces the flux to half of its maximum value. Therefore the correct answer is 15 N m² C⁻¹.

The electric flux through a surface is given by: The square has side: Therefore, Since the plane is parallel to the yz plane, its area vector is along the x-axis. The electric field is also along the x-axis. Hence, and Unit verification: Therefore the correct answer is 30 N m² C⁻¹.

The area associated with a surface element is represented by an area vector whose direction is normal to the surface. Electric flux is: or where θ is the angle between the electric field and the area vector. As the orientation of the surface changes, the value of changes. Consequently, the effective projected area normal to the electric field changes. Therefore, the number of electric field lines crossing the area element also changes with orientation. Statements A, B and D are standard NCERT facts. Statement C is incorrect because electric flux explicitly depends on orientation.

For every closed surface, the area vector is defined according to the outward normal convention. In a cylindrical Gaussian surface, the curved surface consists of infinitely many small area elements. At each point, the area vector is perpendicular to the surface and directed outward from the enclosed volume. This convention is essential because Gauss's law is written as: where always represents the outward area vector. The direction is therefore not arbitrary and is not parallel to the cylinder axis. Using inward normals would reverse the sign of the calculated flux. Hence the correct answer is Option C.

Consider a closed cylinder whose axis is parallel to a uniform electric field. For the flat face whose outward normal points along the electric field: Thus, which is positive. For the opposite face, the outward normal points opposite to the electric field. Hence, which is negative. For the curved surface, the electric field is perpendicular to the outward normal at every point. Therefore, The total flux becomes: Hence the fluxes through the two flat faces are positive and negative respectively.

For a closed surface, the total electric flux is obtained by dividing the surface into very small area elements and summing the contribution from each element. In the limit when the area elements become infinitesimally small, the summation becomes a surface integral: The electric field may vary from point to point over the surface, but over an infinitesimally small element , it can be treated as approximately constant. Therefore statements 1, 2 and 3 are correct. Statement 4 is incorrect because Gauss's law and surface integration remain valid even when the electric field varies across the surface.

According to Gauss's law, A dipole contains equal positive and negative charges. Hence its net enclosed charge is zero and the total flux is zero. A cube enclosing charge has total flux . An uncharged box in a uniform electric field has as much flux entering as leaving, giving zero net flux. A single positive enclosed charge produces a non-zero outward flux. Therefore: 1 → b 2 → a 3 → c 4 → d Hence Option A is correct.

Gauss's law states that the net electric flux through a closed surface depends only on the net charge enclosed by that surface: The electric field appearing in the integral, however, is the resultant field produced by all charges present, both inside and outside the Gaussian surface. Although external charges contribute to the field at points on the surface, their net contribution to the total flux through the closed surface is zero. Therefore statements 1 and 3 are correct.

A Gaussian surface is an imaginary closed surface chosen for applying Gauss's law. It may be spherical, cylindrical, cubical or any other convenient closed shape. However, it should not pass through a discrete point charge because the electric field becomes singular at the location of a point charge, making the flux calculation problematic. The shape is chosen mainly to exploit symmetry and simplify the electric field calculation.

Gauss's law is always valid, but it becomes useful for calculating electric fields only when symmetry is present. A Gaussian surface should be chosen so that the electric field has the same magnitude everywhere on the relevant part of the surface and is either normal or parallel to the area vector. This allows the electric field term to be taken outside the integral and simplifies the evaluation of flux. Such symmetry occurs for spherical, cylindrical and planar charge distributions.

For an infinitely long line charge:

E=λ2πε0r

Using

14πε0=9×109

we get

12πε0=18×109

Given:

E=9×104  N/C

r=2 cm=0.02 m

Substituting,

9×104=λ(18×109)0.02

λ=9×104)(0.0218×109

λ=10−7  C/m

Therefore the linear charge density is:

10−7  C/m

Diameter:

d=2.4 m

Radius:

r=1.2 m

Surface area:

A=4πr2

A=4π(1.2)2

A≈18.1 m2

Surface charge density:

σ=80×10−6 C/m2

Total charge:

q=σA

q=(80×10−6)(18.1)

q≈1.45×10−3 C

Hence the total charge is approximately:

1.45×10−3 C

In electrostatics, continuous charge distributions are described using charge density. The volume element must be sufficiently small compared with the dimensions of the body so that local variations can be studied. However, must still contain an enormous number of microscopic charged particles such as electrons, ions and atoms. This permits averaging and smoothing of the charge distribution. Therefore, is macroscopically small but microscopically large enough to contain many charged constituents. Hence the correct answer is: Macroscopic, Microscopic.

The macroscopic smoothing concept replaces the actual discrete arrangement of charges with an equivalent continuous charge distribution. This approach is justified because the number of microscopic charges is extremely large. Instead of tracking each electron and ion separately, one uses averaged quantities such as linear, surface and volume charge densities. Thus the method ignores the discrete molecular constitution and the empty spaces between particles. It is analogous to how mass density is treated in mechanics. Therefore Option B is incorrect because continuous charge density does not fully account for every microscopic detail.

An infinite line charge exhibits cylindrical symmetry. Therefore, a cylindrical Gaussian surface is chosen while applying Gauss's law.

For a positive linear charge density λ, the electric field points radially outward from the wire.

For a negative linear charge density λ, the electric field points radially inward toward the wire.

Applying Gauss's law:

E(2πrl)=λlε0

which gives

E=λ2πε0r

Thus the field varies inversely with radial distance r.

Hence:

1 → b

2 → c

3 → a

4 → d

For an infinite line charge having linear charge density , a cylindrical Gaussian surface of radius and length is chosen. By symmetry, the electric field is radial and constant over the curved surface. Applying Gauss's law: Cancelling from both sides: Thus, the electric field depends only on the radial distance and is completely independent of the chosen length of the Gaussian cylinder. This is an important result obtained from cylindrical symmetry and Gauss's law.

For an infinite uniformly charged plane sheet, the electric field is perpendicular to the sheet and has the same magnitude on both sides. A cylindrical pillbox or a parallelepiped Gaussian surface may be chosen. Since the electric field is parallel to the curved side surface, the flux through the side faces is zero. Flux passes only through the two flat faces. Therefore total flux is: Thus statement 3 is incorrect because the total flux is not . Statements 1, 2 and 4 are correct. Using Gauss's law ultimately gives:

Each infinite sheet produces a field of magnitude: Outside the pair of sheets, the fields due to the two sheets are equal and opposite, so they cancel. Hence the electric field outside is zero. Between the sheets, both fields act in the same direction and therefore add: Since the sheets are infinite, the field does not depend on position and remains uniform throughout the region between them. Therefore statements 1, 2 and 4 are correct.

For a uniformly charged thin spherical shell of radius , Gauss's law is applied using a concentric spherical Gaussian surface of radius . Since the electric field is constant over the Gaussian surface: Hence, This is exactly the same expression as the electric field due to a point charge located at the centre. Therefore, for all external points, the entire shell behaves as though its total charge were concentrated at its centre.

For a uniformly charged spherical shell, Gauss's law shows that any Gaussian surface drawn completely inside the shell encloses no charge. Therefore: which leads to everywhere inside the shell. This remarkable result is valid only because the electrostatic force obeys an inverse-square law. If Coulomb's law had any other dependence such as or , the exact cancellation producing zero electric field inside the shell would not occur. Thus experimental verification of the zero interior field provides strong evidence for the dependence of Coulomb's law.