CUET UG Physics Booster Test – 3 Point Charges and Dipoles

While deriving the electrostatic potential due to a point charge, an intermediate point P' is considered at a distance r' from the source charge. The position vector of this point is represented by r'. According to Coulomb's law, the electric force acting on a unit positive test charge at P' is As the test charge is brought from infinity toward the charge, the radial displacement element is directed inward, making Δr' negative. This sign plays an important role in obtaining the correct expression for work and potential. Since electrostatic forces are conservative, the total work done depends only on the initial and final positions and not on the path taken. Therefore statement 4 is incorrect. The first three statements correctly describe the derivation of potential from Coulomb's law.

For a point charge, This expression shows that potential is inversely proportional to distance. If the distance is doubled from r to 2r, Similarly, if the distance is reduced from r to r/2, Thus doubling the distance halves the potential, while halving the distance doubles the potential. This inverse relationship is a fundamental characteristic of point-charge potentials and helps explain the behavior of equipotential surfaces around charged particles.

When both the source charge and test charge are positive, the electrostatic force between them is repulsive. Bringing the test charge closer to the source charge requires work to be done against this repulsive force. Therefore, an external agent must supply positive work. Under quasi-static conditions, this work is stored as electrostatic potential energy. As a result, the potential energy of the system increases when the charges are brought closer together. Hence This increase in potential energy reflects the energy stored in the electric field configuration. The concept is analogous to compressing a spring, where external work increases the stored potential energy.

A positive test charge brought toward a negative source charge experiences an attractive electrostatic force. Since the force and displacement are in the same direction, the electric field performs positive work. The work done by the field results in a decrease in the potential energy of the system. Therefore, This behavior is analogous to an object falling under gravity, where potential energy decreases as the object moves in the direction of the force. The external force required for slow motion acts opposite to the displacement and performs negative work. Thus the decrease in potential energy is a direct consequence of the attractive interaction between opposite charges.

For a point charge, and Therefore, statements 1 and 3 are correct. At r = 1 m, both expressions reduce numerically to Q when constants are ignored, making statement 4 correct. Statement 2 is incorrect because electric field represents force per unit charge whereas electric potential represents work done per unit charge. Electric field is a vector quantity while electric potential is a scalar quantity.

For a point charge, the electric potential varies as and the electric field varies as For an electric dipole at large distances, and Therefore, the potential due to a dipole does not decrease as 1/r. Instead, it decreases more rapidly as 1/r². Similarly, the dipole electric field decreases as 1/r³, which is faster than the 1/r² dependence of a point charge field. This rapid fall-off explains why dipole effects become negligible at large distances more quickly than the effects of isolated charges.

The magnitude of electric dipole moment is given by where q is the magnitude of either charge and 2a is the separation between the charges. Given: Substituting, Therefore, the magnitude of the dipole moment is The dipole moment measures the strength of charge separation and is directed from the negative charge toward the positive charge.

The exact electrostatic potential due to an electric dipole is obtained by applying the superposition principle. Since potential is a scalar quantity, the total potential is simply the algebraic sum of the potentials due to the individual charges: This equation is exact and does not require any approximation. Approximations are introduced only later when the condition r >> a is applied to derive the simplified dipole potential formula. The superposition principle assumes that each charge contributes independently to the total potential. Because potential is scalar, contributions can be added algebraically without vector analysis. This property makes potential calculations significantly easier than electric field calculations.

In dipole calculations, the observation point is often located far from the dipole compared to the separation between its charges. Mathematically, this condition is written as When this condition is satisfied, the ratio a/r becomes very small. As a result, higher-order terms such as (a/r)², (a/r)³ and beyond contribute negligibly and can be ignored. This simplification allows the use of binomial expansion and leads to the standard approximate expressions for dipole potential and electric field. The approximation greatly simplifies calculations while maintaining excellent accuracy for distant points.

For points far from the dipole, the terms involving a/r are very small. Applying the binomial approximation gives and These approximations are crucial in deriving the standard dipole potential formula. The expressions simplify complicated distance terms and make analytical derivations possible while preserving the dominant physical behavior of the dipole field.

For a physical dipole observed at distances much larger than its size (r >> a), the electric potential is given by This expression shows that the potential depends on the dipole moment p, the angle θ and the inverse square of distance. When θ = π/2, cosθ = 0 and therefore the potential becomes zero. This corresponds to the equatorial plane of the dipole. For θ = 0, cosθ = +1 and the potential is positive. For θ = π, cosθ = −1 and the potential becomes negative. Hence the potential changes sign across the dipole axis. Statement 3 is incorrect because a dipole potential varies as 1/r², whereas the potential due to a point charge varies as 1/r.

The vector representation of dipole potential is The dot product can be written as which gives the familiar scalar expression for dipole potential. The vector r̂ is the unit vector directed from the dipole toward the observation point. The formula clearly shows that potential varies inversely as r² and not as r³. The 1/r³ dependence is associated with the electric field of a dipole rather than its potential. For an ideal point dipole, this expression is exact. Therefore option C is the incorrect statement.

For a dipole, the potential at a distant point is At θ = 0, therefore which is the maximum positive potential on the axis. At θ = π, therefore which is the minimum negative potential on the axis. Hence statements 1 and 2 are correct. Statement 3 is correct because the axial potential is represented by . Statement 4 is also correct because the sign of the potential changes as the value of cosθ changes from +1 to −1 when moving from θ = 0 to θ = π.

For every point on the equatorial plane of a dipole, Therefore, and the dipole potential becomes zero. Since all points on the plane have the same potential, the plane is an equipotential surface. When a test charge moves along an equipotential surface, the potential difference is zero and therefore no work is done by the electric field. The potential energy of the charge remains constant throughout the motion. However, the electric field on the equatorial plane is not zero. It has a finite magnitude and is directed opposite to the dipole moment vector. Hence statement 4 is incorrect.

The dipole potential is given by The potential depends only on the radial distance r and the angle θ between the dipole moment vector and the position vector. If the position vector is rotated around the dipole moment axis while keeping θ fixed, neither r nor θ changes. Therefore the value of the potential remains exactly the same at all such points. The collection of these points forms a conical surface around the dipole axis. This property demonstrates the axial symmetry of dipole potential. Unlike a point charge, which possesses spherical symmetry, a dipole possesses rotational symmetry only about its dipole axis.

An electric dipole consists of two equal and opposite charges separated by a small distance. At very large distances, the observation point is nearly equidistant from both charges. Consequently, the potential due to the positive charge and the potential due to the negative charge almost cancel each other. Because the net charge of the dipole is zero, the dominant 1/r term disappears. The remaining non-zero contribution arises from the separation of charges and is proportional to 1/r². This is why the dipole potential decreases faster than the potential due to a single point charge. The cancellation of the leading term is the fundamental mathematical and physical reason behind the inverse-square dependence of dipole potential.

Let the positive charge +3 × 10⁻⁸ C be at x = 0 and the negative charge −2 × 10⁻⁸ C be at x = 15 cm. For zero potential, Cancelling common factors, Cross multiplying, Thus, the zero-potential point lies 9 cm from the positive charge. This result follows directly from the principle of superposition of electrostatic potentials.

For a system of point charges, the potential due to an individual charge is where is the distance between the charge and the observation point P. Therefore, the potential V₁ due to charge q₁ contains the denominator r₁P, while the potential V₂ due to charge q₂ contains the denominator r₂P. The notation clearly identifies the source charge and the observation point. This convention is used extensively in NCERT when applying the superposition principle to multiple-charge systems.

For a continuous charge distribution, the charge is spread throughout a region of space rather than concentrated at discrete points. The distribution is divided into infinitesimally small volume elements Δv. If the volume charge density is ρ, then the charge associated with an element is The potential due to each element is calculated separately and then integrated over the entire distribution. Thus, the discrete summation used for point charges is replaced by a continuous integral. Statement 1 is incorrect because the volume elements must be infinitesimally small, not macroscopically large. Statements 2, 3 and 4 correctly describe the method used in NCERT for calculating potential due to continuous charge distributions.

According to the shell theorem, the electric field inside a uniformly charged spherical shell is zero at every point. Since electric field is related to the change in potential, a zero electric field implies that the potential remains constant throughout the interior of the shell. Therefore, when a test charge moves from one point to another inside the shell, there is no potential difference between the two points. Since and the work done is zero. Consequently, the potential energy of the test charge remains unchanged during motion inside the shell. This remarkable property is one of the most important results associated with spherical symmetry in electrostatics.