CUET UG Physics Booster Test – 3 Flux and Gauss’s Law

Electric flux is defined as: or where is the angle between the electric field vector and the area vector . The area vector is always directed normal to the surface. Therefore, the angle used in the dot product is not the angle between the electric field and the plane itself. If the plane is perpendicular to the electric field, the area vector becomes parallel to the electric field and the flux is maximum. Electric flux is a scalar quantity because it is obtained through a dot product. Thus option D is the incorrect statement and is therefore the correct answer.

Using the formula: For case 1, , therefore . For case 2, , therefore . For case 3, , For case 4, , Hence: 1 → c 2 → a 3 → b 4 → d Therefore Option C is correct.

For any open surface, two opposite normal directions can be drawn. Therefore, there are two possible area vectors. For a closed surface, ambiguity is removed by adopting a standard convention. In electrostatics and Gauss's law, the outward normal is always chosen. This convention ensures consistency in calculating electric flux and interpreting positive and negative flux values. Hence, the correct completion is "Two, Outward."

For a closed cube, area vectors are directed outward from each face. The top face lies perpendicular to the positive z-axis, so its outward normal is . The bottom face lies perpendicular to the negative z-axis, so its outward normal is . These directions follow directly from the outward normal convention used for closed surfaces in flux calculations and Gauss's law.

According to Gauss's law: If the net outward flux is zero, the net enclosed charge must be zero. However, this does not mean that there are no charges inside. Equal positive and negative charges may be present whose algebraic sum is zero. The electric field at the surface may still be non-zero due to internal charge arrangements or external charges. Therefore statements 1 and 2 are correct, while statements 3 and 4 are incorrect.

The cube extends from to . Since the electric field exists only along the x-direction, flux through the four side faces is zero. Flux through left face: Flux through right face: Therefore, Hence the correct answer is Option B.

Imagine the given square as one face of a cube of side 10 cm. Since the charge is located 5 cm above the centre of the square, it lies exactly at the centre of the cube. By Gauss's law, Because of symmetry, all six faces of the cube receive equal flux. Thus, Hence Option B is correct.

From Gauss's law: Therefore, Substituting values, The negative sign indicates that the enclosed charge is negative.

A Gaussian surface may pass through a continuous charge distribution because the charge density remains finite and mathematically manageable. Gauss's law fundamentally follows from Coulomb's inverse-square law. If nature did not obey the inverse-square law, Gauss's law would not hold in its familiar form. At the exact location of a discrete point charge, the electric field becomes infinite and therefore is not well defined. Thus statement 3 is incorrect. Hence statements 1, 2 and 4 are correct.

Gauss's law, is universally valid for all charge distributions, whether symmetric or asymmetric. However, for highly symmetric charge distributions such as spherical, cylindrical or planar systems, the electric field can be taken outside the integral and calculated easily. In asymmetric configurations, although the law remains true, extracting the value of from the integral becomes difficult because the field varies in magnitude and direction over the surface. Therefore Gauss's law remains valid but loses its computational simplicity.

For an infinite straight wire, cylindrical symmetry ensures that the electric field is purely radial and depends only on the perpendicular distance from the wire. However, a finite straight wire does not possess perfect cylindrical symmetry. Near the ends of the wire, electric field lines bend and are no longer strictly radial. Therefore, the electric field cannot be treated as having constant magnitude over a cylindrical Gaussian surface, making direct extraction of E from Gauss's law difficult. Gauss's law remains valid because it is a fundamental law of electrostatics. However, due to reduced symmetry, it becomes less useful for directly calculating the electric field. Only near the midpoint of a very long wire does the field approximate the radial pattern of an infinite wire. Hence, statement D is incorrect.

NCERT defines charge densities according to the dimensional nature of charge distribution. For a linear distribution: For a surface distribution: For a volume distribution: A continuous charge distribution is an idealized description where charge is assumed to be smoothly spread over a region containing a very large number of microscopic charged constituents. This approximation simplifies electrostatic calculations and is widely used in Gauss's law applications. Therefore: 1-b, 2-a, 3-c, 4-d.

For a neutral atom: Hence, Therefore, The negative sign indicates the electron cloud surrounding the positively charged nucleus. Thus, Option B is correct.

For a Gaussian sphere of radius r: Positive enclosed charge: Negative enclosed charge: Net enclosed charge: Applying Gauss's law, After simplification, Thus Option A is correct.

For an infinite line charge, Gauss's law gives: The magnitude is inversely proportional to radial distance r. If is negative, the electric field lines point towards the wire because electric field lines terminate on negative charges. Hence the direction is radially inward. Therefore: Direction → Inward Variation → Inversely proportional to r Thus Option D is correct.

An infinite straight uniformly charged wire possesses cylindrical symmetry. Therefore, rotating any radial vector around the wire axis produces points that are physically equivalent. Since these points are equivalent, the electric field magnitude must remain the same at all such points. During the derivation of the electric field, charge elements symmetrically located on opposite sides of the observation point produce transverse components that cancel each other. Only radial components survive and add together. The electric field is directed radially outward for positive charge density and radially inward for negative charge density. Hence statement 3 is incorrect because the field is not parallel to the wire. Therefore statements 1, 2 and 4 are correct.

The expression is obtained assuming an infinite uniformly charged plane sheet. Real sheets are finite and therefore exhibit edge effects near their boundaries. However, when the dimensions of the sheet are much larger than the observation distance and the observation point lies near the central region, the influence of the edges becomes negligible. Under these conditions, the electric field closely approximates the value predicted for an infinite sheet. Hence the formula is practically applicable in the middle regions of a sufficiently large sheet.

For an infinite uniformly charged plane sheet placed in the yz-plane, symmetry demands that the electric field be perpendicular to the sheet. Therefore, the field must be directed along the positive or negative x-axis. Since the sheet is infinite, there is no preferred position along the y or z directions, making the field magnitude independent of y and z coordinates. Furthermore, the magnitude is also independent of x. Using a Gaussian pillbox, flux passes through two parallel faces, giving total flux: Hence statements 1, 3 and 4 are correct while statement 2 is incorrect.

Consider a Gaussian sphere of radius surrounding the charged solid sphere. By spherical symmetry, the electric field has the same magnitude everywhere on the Gaussian surface and is directed radially outward. Applying Gauss's law: Thus, This is exactly the electric field produced by a point charge q placed at the center of the sphere. Therefore, a uniformly charged solid sphere and a uniformly charged thin spherical shell produce identical external electric fields.

The remarkable result that the electric field inside a uniformly charged spherical shell is exactly zero depends critically on Coulomb's inverse-square law: This specific dependence ensures perfect cancellation of electric field contributions from all parts of the shell. If the force law were instead proportional to , the cancellation would no longer occur exactly. Contributions from different regions of the shell would not balance each other perfectly, leading to a non-zero electric field inside the shell. Moreover, the field would vary from point to point depending on position within the shell. This fact highlights the deep connection between Gauss's law and the inverse-square nature of electrostatic interactions.