CUET UG Physics Booster Test – 2 Point Charges and Dipoles

Electrostatic potential is defined as the work done by an external force in bringing a unit positive test charge from infinity to a given point. Therefore, the integration is performed from r' = ∞ to r' = r. During the derivation, the displacement element is directed inward toward the charge, producing a negative radial increment. This leads to the appearance of a negative sign that ultimately yields a positive value for the work done by the external agent. The charge is assumed to move infinitesimally slowly so that its kinetic energy remains essentially constant throughout the process. Since electrostatic forces are conservative, the work done depends only on the initial and final positions and not on the path followed. Therefore, statement 4 is incorrect. The first three statements correctly describe the derivation and physical meaning of electrostatic potential.

The potential due to a point charge is given by V = kQ/r Rearranging, r = kQ/V Substituting the values: r = (9 × 10⁹ × 4 × 10⁻⁷)/(4 × 10⁴) r = (36 × 10²)/(4 × 10⁴) r = 0.09 m Converting to centimetres: 0.09 m = 9 cm Therefore, the required distance is 9 cm. This numerical problem directly applies the inverse relationship between potential and distance. As the distance from the charge increases, the potential decreases. Such calculations are frequently used in electrostatics to determine the location of points having a specified potential.

When both the source charge and test charge are positive, the electrostatic force between them is repulsive. Therefore, to bring the test charge closer to the source charge, an external agent must push it against the repulsive electric force. The work done by the external force is positive because energy must be supplied to the system. Under the assumption of infinitesimally slow motion, the kinetic energy remains unchanged and the supplied work is stored entirely as electrostatic potential energy. Thus statements 1 and 3 are correct. Statement 2 is incorrect because the external work is positive, not negative. Statement 4 is also incorrect because the kinetic energy does not increase significantly during a quasi-static process.

A positive test charge experiences an attractive force toward a negative source charge. If the charge is brought slowly without acceleration, the external force must oppose the electrostatic attraction. Therefore, the external force acts opposite to the displacement, not in the same direction. This makes statement B incorrect. The work done by the external force is negative because the electric field assists the motion. The potential at any point due to a negative source charge is also negative, as given by V = (1/4πε₀)(Q/r) where Q is negative. Thus statements A, C and D are correct. The only incorrect statement is B.

For a point charge, electric potential varies as V ∝ 1/r whereas electric field varies as E ∝ 1/r² Since the square of the distance appears in the denominator for electric field, its value decreases much more rapidly than potential as distance increases. Graphically, the 1/r² curve drops sharply near the source and approaches zero more quickly than the 1/r curve. This distinction is important because potential remains significant at larger distances while electric field becomes comparatively weaker. The difference in their rates of decrease helps explain why potential and field exhibit different spatial behaviors around charged objects.

For a point charge, the electric field is given by E = (1/4πε₀)(Q/r²) whereas the electric potential is V = (1/4πε₀)(Q/r) The inverse-square dependence causes the electric field to decrease more rapidly than the potential as distance increases. Therefore, the electric field curve falls more sharply while the potential curve decreases more gradually. This difference becomes particularly noticeable at large distances from the charge. The potential remains appreciable even where the field has become relatively weak. This distinction is important in understanding electrostatic interactions and comparing the spatial behavior of field and potential. NCERT emphasizes that potential and field are related but possess different distance dependences due to the mathematical relationship between them.

An electric dipole consists of two equal and opposite charges separated by a small distance. If the charges are +q and −q and the separation is 2a, the dipole moment magnitude is p = q(2a) Therefore, statements 1 and 4 are correct. Since the charges are equal and opposite, the algebraic sum of charges is zero. Thus statement 2 is incorrect. According to NCERT, the dipole moment vector is directed from the negative charge toward the positive charge. Hence statement 3 is incorrect because it reverses the direction. The dipole is electrically neutral but possesses a non-zero dipole moment, which determines its interaction with external electric fields and contributes to its unique potential distribution.

According to the superposition principle, the total potential at a point equals the algebraic sum of the potentials due to individual charges. For a dipole consisting of charges +q and −q, V = (1/4πε₀)(q/r₁) + (1/4πε₀)(−q/r₂) Combining these terms gives V = (1/4πε₀)[q/r₁ − q/r₂] The negative sign arises because one charge is negative. Since potential is a scalar quantity, algebraic addition is sufficient and vector addition is not required. This expression forms the basis for deriving the approximate dipole potential formula applicable when the observation point is far from the dipole. The superposition principle is one of the most important concepts in electrostatics because it allows complex charge systems to be analyzed using simple additions.

While deriving the potential due to an electric dipole, the observation point is located at a distance r from the dipole center. The distance from the positive charge to the observation point is represented by r₁. Applying the law of cosines to the relevant triangle gives r₁² = r² + a² − 2ar cosθ Similarly, r₂² = r² + a² + 2ar cosθ These geometric relations are essential because the exact dipole potential depends on the distances from both charges. The expressions are later simplified using binomial expansion when the observation point is far from the dipole. Understanding these geometric relations helps in deriving the standard dipole potential formula and explains the dependence of potential on both distance and orientation.

In the derivation of dipole potential, the observation point is assumed to be far away from the dipole. This means r >> a Under this condition, the quantity a/r becomes very small and binomial expansion can be applied. Only first-order terms in a/r are retained because higher-order terms contribute negligibly. Using this approximation, 1/r₁ ≈ (1/r)[1 + (a/r)cosθ] and a similar expression is obtained for 1/r₂. These approximations simplify the exact expression and lead to the well-known dipole potential formula. However, the approximation is not exact for all distances and becomes increasingly accurate only when r is much greater than a. Therefore statement 4 is incorrect.

The standard expression for the potential due to an electric dipole at a distant point is This expression is derived under the approximation that the observation point is very far from the dipole compared to its size, that is, r >> a. The formula clearly shows that the potential depends directly on the dipole moment magnitude p. Therefore, a larger dipole moment produces a larger potential at the same location. The expression also contains the factor cosθ, indicating that the potential depends on the orientation of the observation point relative to the dipole axis. Since r² appears in the denominator, the potential decreases as the inverse square of distance. Thus statements A, B and C are correct, whereas statement D is incorrect because the dipole moment magnitude explicitly appears in the formula.

The vector representation of dipole potential is written as where p is the dipole moment vector and r̂ is the unit vector directed from the dipole toward the observation point. The dot product can be expanded as which immediately produces the scalar form of the dipole potential. This representation provides a compact mathematical description and clearly shows the directional dependence of the potential. The formula is derived using the approximation r >> a and therefore applies to points located far from the dipole. It is not valid only when r = a. The vector form highlights the importance of orientation in dipole fields and potentials.

For a point on the dipole axis, Since θ = 0, Substituting the given values: Therefore, The positive sign indicates that the observation point lies on the positive axial side of the dipole. This calculation demonstrates how dipole potential depends on both distance and orientation. On the axis, the potential attains its maximum positive or negative value depending on the side of the dipole being considered.

The dipole potential is given by For points lying on the equatorial plane, and Hence, This means that every point on the equatorial plane has zero potential, the same value as the chosen reference point at infinity. Therefore the equatorial plane forms an equipotential surface. Since the potential is zero everywhere on this plane, no work is required to move a test charge between any two points on it. This property is an important consequence of the symmetry of an electric dipole.

The potential due to a point charge is Since only the distance r appears in the expression, the potential is the same in all directions. Therefore it is spherically symmetric. In contrast, the dipole potential is which depends on the angle θ. Because of this directional dependence, the dipole potential is not spherically symmetric. Instead, it possesses axial symmetry about the dipole moment vector p. Rotating the system around the dipole axis does not change the potential, whereas changing θ alters the value of the potential. These symmetry differences are fundamental in electrostatics and help distinguish point charges from dipoles.

The potential due to a point charge is given by Therefore, point charge potential falls off as 1/r. The electric field due to a point charge is and thus decreases as 1/r². For an electric dipole at large distances, the potential is which falls as 1/r². The dipole electric field decreases even more rapidly, varying as 1/r³. These different rates of decrease explain why the influence of a dipole becomes negligible more quickly than that of a single isolated charge. Hence statements 1 and 3 are correct, while statements 2 and 4 are incorrect.

The electrostatic potential due to a system of charges is obtained by applying the superposition principle. Since potential is a scalar quantity, individual potentials are added algebraically rather than vectorially. Therefore, Substituting the given values, The negative sign associated with V₂ must be retained during addition because potential is a signed scalar quantity. This simple numerical example illustrates the power of the superposition principle, which allows the potential due to a complex system of charges to be determined by adding the contributions from each charge independently.

For a system of point charges, the total potential at a point P is where qᵢ is the ith charge and rᵢP represents the distance between that charge and the observation point P. Each charge contributes independently to the total potential. Therefore, r₁P specifically denotes the distance between the first charge q₁ and the observation point P. This distance determines the magnitude of the contribution of q₁ to the total potential. The quantity does not refer to the origin, nor does it represent the distance between different charges.

For a continuous charge distribution, the charge is spread throughout a region rather than concentrated at discrete points. To calculate the potential, the distribution is divided into many small volume elements Δv. If ρ is the volume charge density, then the charge contained in a small element is Each element contributes a small potential to the observation point. The total potential is obtained by summing the contributions of all such elements and then taking the continuous limit, resulting in an integral. This method is used extensively in electrostatics to determine the potential due to charged rods, discs, spheres and other continuous charge distributions. Since ρ represents the charge density of the distribution, it cannot be assumed to be zero everywhere.

According to the shell theorem discussed in NCERT, a uniformly charged spherical shell behaves like a point charge for points outside the shell. Therefore, for r ≥ R. Inside the shell, the electric field is zero everywhere. Since electric field is the negative gradient of potential, a zero electric field implies that the potential remains constant throughout the interior region. The value of this constant potential is equal to the potential at the surface of the shell. Therefore, the statement that the potential inside varies as 1/r is incorrect. Such a variation applies only outside the shell where the shell behaves like a point charge.