CUET UG Physics Booster Test - 3 Equipotentials and Energy -

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CUET UG Physics Booster Test - 3 Equipotentials and Energy

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Physics Booster Test - 3 Equipotentials and Energy

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QUESTION 1 OF 20

Constant Potential Value Across Entire Surface implies:
Statements:
1. The normal component of the electric field dictates the potential change between different surfaces.
2. It mathematically disproves Gauss's law.
3. The change in potential (dV) is exactly zero for any path taken exclusively along the surface.
4. The electric field magnitude (E) must be exactly zero everywhere on the surface.

Answer Review

1 Constant Potential Value Across Entire Surface implies:
Statements:
1. The normal component of the electric field dictates the potential change between different surfaces.
2. It mathematically disproves Gauss's law.
3. The change in potential (dV) is exactly zero for any path taken exclusively along the surface.
4. The electric field magnitude (E) must be exactly zero everywhere on the surface.

A 1 and 3 are correct

B 2 and 4 are correct

C 1 and 2 are correct

D 3 and 4 are correct

💡 Short Explanation

�� Potential remains constant on an equipotential surface. �� No potential change occurs along the surface. �� Electric field causes potential change only in the normal direction.

✅ Correct Answer Explanation

An equipotential surface is defined as a surface on which the electric potential has the same value at every point. Therefore, any displacement entirely along the surface produces no change in potential, making . The electric field is always perpendicular to an equipotential surface and is responsible for changes in potential between neighboring equipotential surfaces. Hence, statement 1 is correct because the normal component of the field determines how rapidly potential changes from one surface to another. Statement 3 is also correct because the potential remains constant for any path lying completely on the surface. Statement 2 is incorrect because equipotential surfaces are fully consistent with Gauss's law. Statement 4 is incorrect because the electric field can be non-zero while remaining perpendicular to the surface.

❓ Why Other Options Are Incorrect

�� Option B → Statements 2 and 4 are incorrect.
�� Option C → Statement 2 is incorrect.
�� Option D → Statement 4 is incorrect.

🎯 Strategy

Used – Concept Application

Application
Use the definition of an equipotential surface and the relationship between electric field and potential.
Final Logic
Potential remains constant along the surface and changes only in the normal direction.

🧠 Memory Trick

Normal Direction → Potential Change

📚 Class: XII | Book: NCERT Physics Part I | Chapter: Electrostatic Potential and Capacitance | Topic: Defining Equipotential Surfaces | Subtopic: Constant Potential Value Across Entire Surface

2 If non-zero work was inherently required to move a charge freely along an equipotential surface, this would mathematically contradict:

A Coulomb's law of interaction

B The defining properties of an equipotential surface

C The basic superposition principle

D Gauss's law for enclosed charges

💡 Short Explanation

�� Equipotential surfaces have constant potential. �� Work done depends on potential difference. �� No potential difference means no work.

✅ Correct Answer Explanation

The work done in moving a charge q between two points is For an equipotential surface, Therefore, If non-zero work were required to move a charge along the surface, then a non-zero potential difference would exist between points on the same surface. This would directly violate the definition of an equipotential surface. Thus, the assumption of non-zero work contradicts the fundamental properties of equipotential surfaces.

❓ Why Other Options Are Incorrect

�� Option A → Coulomb's law is unrelated to the definition of equipotential surfaces.
�� Option C → Superposition principle is unaffected.
�� Option D → Gauss's law is not violated.

🎯 Strategy

Used – Logical Analysis

Application
Apply the work-potential relationship.
Final Logic
Non-zero work implies non-zero potential difference, which is impossible on an equipotential surface.

🧠 Memory Trick

Equipotential → Equal Potential → Zero Work

📚 Class: XII | Book: NCERT Physics Part I | Chapter: Electrostatic Potential and Capacitance | Topic: Defining Equipotential Surfaces | Subtopic: Zero Work for Motion Along Surface

3 Incorrect statement regarding geometric shapes of equipotential surfaces for a single point charge q:

A The potential V is mathematically constant if radius r is constant.

B Field lines point outward or inward depending on whether q is positive or negative.

C The electric field lines are strictly radial.

D The surfaces are modeled as extending concentric cylinders.

💡 Short Explanation

�� Equipotential surfaces around a point charge are spherical. �� Electric field lines are radial. �� Cylindrical symmetry does not apply.

✅ Correct Answer Explanation

For a point charge, The potential depends only on the radial distance r from the charge. Therefore, all points at the same distance possess the same potential and form a spherical surface. Hence, the equipotential surfaces are concentric spheres centered on the charge. Electric field lines originate radially outward from a positive charge and terminate radially inward toward a negative charge. Statement B is incorrect because concentric cylinders are not the equipotential surfaces of a point charge.

❓ Why Other Options Are Incorrect

�� Option A → Correct because constant r gives constant potential.
�� Option B → Correct description for positive and negative charges.
�� Option C → Correct description of field lines.

🎯 Strategy

Used – NCERT Recall

Application
Recall the geometry of the electric field and potential around a point charge.
Final Logic
Point charges produce spherical equipotential surfaces.

🧠 Memory Trick

Point Charge → Spheres

📚 Class: XII | Book: NCERT Physics Part I | Chapter: Electrostatic Potential and Capacitance | Topic: Geometric Shapes of Surfaces | Subtopic: Concentric Spheres for Isolated Point Charges

4 Statements:
1. The field magnitude decreases exponentially with x.
2. The work done moving a test charge entirely within the y-z plane is exactly zero.
3. The equipotential surfaces are aligned parallel to the x-axis.
4. The equipotential surfaces are aligned parallel to the y-z plane.

A 2 and 3 are correct

B 1 and 3 are correct

C 1 and 4 are correct

D 2 and 4 are correct

💡 Short Explanation

�� Uniform field points along x-axis. �� Equipotential surfaces are perpendicular to the field. �� Motion within the y-z plane requires no work.

✅ Correct Answer Explanation

In a uniform electric field directed along the positive x-axis, equipotential surfaces must be perpendicular to the field lines. Therefore, they are planes parallel to the y-z plane. Since all points on a given equipotential plane have the same potential, moving a charge entirely within that plane requires no work. Statement 3 is incorrect because equipotential surfaces cannot be parallel to the electric field. Statement 1 is incorrect because a uniform electric field has constant magnitude and does not decrease exponentially.

❓ Why Other Options Are Incorrect

�� Option A → Statement 3 is incorrect.
�� Option B → Statements 1 and 3 are incorrect.
�� Option C → Statement 1 is incorrect.

🎯 Strategy

Used – Concept Application

Application
Use the perpendicular relationship between electric field and equipotential surfaces.
Final Logic
Uniform field along x produces equipotential planes parallel to the y-z plane.

🧠 Memory Trick

Field Along x → Equipotential y-z

📚 Class: XII | Book: NCERT Physics Part I | Chapter: Electrostatic Potential and Capacitance | Topic: Geometric Shapes of Surfaces | Subtopic: Parallel Planes for Uniform Electric Fields

5 The formal proof that the electric field must be strictly normal to the equipotential surface at every point is primarily based on:

A The non-existence of isolated magnetic monopoles

B The contradiction that any tangential component would necessitate work to move a charge along the surface

C The fundamental quantization of electric charge

D The continuous volume charge distribution theorem

💡 Short Explanation

�� Equipotential surfaces have constant potential. �� Tangential electric field would perform work. �� Therefore electric field must be normal.

✅ Correct Answer Explanation

An equipotential surface has the same electric potential at every point. Therefore, moving a charge along the surface should require no work. If the electric field had a tangential component along the surface, it would exert a force on the charge and perform work during motion. This would create a potential difference between points on the surface, contradicting the definition of an equipotential surface. Hence, the tangential component of the electric field must be zero, leaving only the normal component. Therefore, electric field lines must always be perpendicular to equipotential surfaces.

❓ Why Other Options Are Incorrect

�� Option A → Magnetic monopoles are unrelated to this proof.
�� Option C → Charge quantization does not determine field orientation.
�� Option D → Volume charge distributions do not provide the fundamental proof.

🎯 Strategy

Used – Logical Analysis

Application
Assume a tangential component exists and examine the consequence.
Final Logic
A tangential field would produce work, violating the equipotential condition.

🧠 Memory Trick

No Tangential Field → No Work

📚 Class: XII | Book: NCERT Physics Part I | Chapter: Electrostatic Potential and Capacitance | Topic: Field and Surface Relationship | Subtopic: Normal Orientation of Field to Surface

6 What physical violation occurs if an electric field E possesses a non-zero tangential component along an equipotential surface in a static situation?

A The free charges on the surface would permanently freeze.

B The equipotential surface would spontaneously become an insulator.

C It would require work to move a charge along the surface, violating the zero-work property.

D The potential would instantly drop to a negative value.

💡 Short Explanation

�� Equipotential surfaces have constant potential. �� Tangential electric field causes work. �� This violates the equipotential condition.

✅ Correct Answer Explanation

An equipotential surface is defined as a surface on which the electric potential remains constant. Therefore, the potential difference between any two points on the surface is zero. If the electric field possessed a tangential component along the surface, a charge moving on the surface would experience a force in the direction of motion. Consequently, work would be done during the displacement. This would imply a non-zero potential difference between points on the same surface, contradicting the definition of an equipotential surface. Hence, in electrostatic equilibrium, the tangential component of the electric field must vanish and the field must remain perpendicular to the equipotential surface.

❓ Why Other Options Are Incorrect

�� Option A → Charges do not freeze because of tangential field components.
�� Option B → Equipotential surfaces do not become insulators.
�� Option D → Potential does not suddenly become negative.

🎯 Strategy

Used – Logical Analysis

Application
Assume a tangential field exists and analyze its consequences.
Final Logic
Tangential field implies work along the surface, which violates the equipotential property.

🧠 Memory Trick

Tangential Field → Work Done → Not Equipotential

📚 Class: XII | Book: NCERT Physics Part I | Chapter: Electrostatic Potential and Capacitance | Topic: Field and Surface Relationship | Subtopic: Non-existence of Tangential Field Components

7 Potential of a dipole perfectly along the equatorial plane:

A Varies as 1/r²

B Is exactly zero

C Is strictly infinite

D Varies as 1/r³

💡 Short Explanation

�� Equatorial plane corresponds to θ = 90°. �� Contributions from +q and −q cancel. �� Net potential becomes zero.

✅ Correct Answer Explanation

For an electric dipole, the potential at a distant point is given by On the equatorial plane, Therefore, and hence Physically, every point on the equatorial plane is equidistant from the positive and negative charges of the dipole. The positive and negative potential contributions therefore cancel exactly. Thus, the entire equatorial plane acts as a zero-potential surface.

❓ Why Other Options Are Incorrect

�� Option A → Potential becomes zero, not .
�� Option C → Potential is finite and zero.
�� Option D → describes neither the equatorial potential nor the dipole potential expression.

🎯 Strategy

Used – Substitution

Application
Substitute into the dipole potential formula.
Final Logic
gives zero potential.

🧠 Memory Trick

Dipole Equator → Potential Zero

📚 Class: XII | Book: NCERT Physics Part I | Chapter: Electrostatic Potential and Capacitance | Topic: Complex Equipotential Patterns | Subtopic: Surface Map for Electric Dipoles

8 Correct statements about complex surface maps:
Statements:
1. For an electric dipole, the potential falls off as 1/r² at large distances.
2. Two identical positive charges have merged, complex equipotential surfaces at larger distances.
3. For a dipole, the equipotential surfaces are perfect concentric spheres around the origin.

A 1 and 2 are correct

B 1 and 3 are correct

C 2 and 3 are correct

D 2 only is correct

💡 Short Explanation

�� Dipole potential falls as 1/r². �� Two-charge systems produce complex patterns. �� Dipole surfaces are not spherical.

✅ Correct Answer Explanation

At large distances, the electric potential due to a dipole is showing a dependence. Thus statement 1 is correct. For two identical positive charges, the equipotential surfaces close to each charge appear nearly spherical, but farther away they merge into more complex shapes due to the combined influence of both charges. Hence statement 2 is correct. Statement 3 is incorrect because concentric spherical equipotential surfaces occur only for a single isolated point charge. A dipole produces angle-dependent equipotential surfaces that are not spherical.

❓ Why Other Options Are Incorrect

�� Option B → Statement 3 is incorrect.
�� Option C → Statement 3 is incorrect.
�� Option D → Statement 1 is also correct.

🎯 Strategy

Used – NCERT Recall

Application
Recall the standard equipotential maps of dipoles and charge pairs.
Final Logic
Dipole potential depends on both distance and angle.

🧠 Memory Trick

Dipole → Shape Changes

📚 Class: XII | Book: NCERT Physics Part I | Chapter: Electrostatic Potential and Capacitance | Topic: Complex Equipotential Patterns | Subtopic: Surface Map for Identical Positive Charges

9 Match the following:

List I	List II
1. Direction of steepest potential decrease	a. E=-δV/δl
2. Electric field direction	b. Direction of electric field
3. Potential gradient relation	c. Rate of potential change per unit length
4. Equipotential surface	d. Surface of constant potential

A 1-b, 2-c, 3-a, 4-d

B 1-a, 2-b, 3-d, 4-c

C 1-c, 2-a, 3-b, 4-d

D 1-b, 2-a, 3-d, 4-c

💡 Short Explanation

�� Electric field points toward decreasing potential. �� Field magnitude is related to potential gradient. �� Equipotential surfaces have constant potential.

✅ Correct Answer Explanation

The electric field always points in the direction of the steepest decrease of potential. Its magnitude is related to the potential gradient through This relation states that electric field equals the rate at which potential decreases with distance. Equipotential surfaces are surfaces on which the potential remains constant. The correct matching follows directly from these definitions and relationships.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect relation assignments.
�� Option C → Direction and gradient concepts are mismatched.
�� Option D → Multiple pairings are incorrect.

🎯 Strategy

Used – Concept Application

Application
Match each physical quantity with its corresponding definition.
Final Logic
Field follows the steepest potential decrease.

🧠 Memory Trick

Field Follows Fall

📚 Class: XII | Book: NCERT Physics Part I | Chapter: Electrostatic Potential and Capacitance | Topic: Field-Potential Mathematical Link | Subtopic: Potential Gradient in Direction of Field

10 Two closely spaced equipotential surfaces A and B possess potentials V and V + dV. If the normal displacement δl is m and V/m, the magnitude of the potential difference is:

A 0.5 V

B V

C 500 V

D 50 V

💡 Short Explanation

�� Use . �� Substitute the given values. �� Compute the potential difference.

✅ Correct Answer Explanation

The magnitude of the electric field is related to potential difference by Given, Therefore, Hence, the potential difference between the two nearby equipotential surfaces is

❓ Why Other Options Are Incorrect

�� Option B → Numerical and unit error.
�� Option C → Distance factor ignored.
�� Option D → Incorrect multiplication.

🎯 Strategy

Used – Substitution

Application
Apply the electric field-potential gradient formula directly.
Final Logic
Potential difference equals field multiplied by normal separation.

🧠 Memory Trick

Voltage Drop = Field × Distance

📚 Class: XII | Book: NCERT Physics Part I | Chapter: Electrostatic Potential and Capacitance | Topic: Field-Potential Mathematical Link | Subtopic: Magnitude as Change in Potential per Unit Length

11 "Electric field is in the direction in which the potential decreases steepest." This directly implies that manually moving a positive test charge opposite to the field direction:

A Continually decreases its internal potential energy

B Performs negative work by the external agency

C Moves the charge to a surface with a higher potential

D Requires exactly zero external work

💡 Short Explanation

�� Electric field points toward decreasing potential. �� Opposite motion means moving toward higher potential. �� Positive charges gain potential energy.

✅ Correct Answer Explanation

The electric field always points in the direction of the steepest decrease in electric potential. Therefore, when a positive test charge is moved opposite to the direction of the electric field, it moves toward regions of higher electric potential. Since the charge is being moved against the electric force, external work must be done on the charge. This work is stored as electrostatic potential energy. Hence, moving opposite to the field direction takes the charge to a surface having a higher potential value.

❓ Why Other Options Are Incorrect

�� Option A → Potential energy increases, not decreases.
�� Option B → External agency performs positive work.
�� Option D → Work is required to move against the field.

🎯 Strategy

Used – Concept Application

Application
Use the relationship between electric field direction and potential gradient.
Final Logic
Opposite to electric field means toward higher potential.

🧠 Memory Trick

Against E → Higher V

📚 Class: XII | Book: NCERT Physics Part I | Chapter: Electrostatic Potential and Capacitance | Topic: Steepest Decrease Principles | Subtopic: Directional Property of the Electric Field

12 Normal displacement and potential principles:
Statements:
1. The change in magnitude of potential per unit normal displacement gives the electric field magnitude.
2. Equipotential surfaces can cross each other if the fields are extremely strong.
3. δV is negative when moving directly in the direction of E.
4. Field lines naturally run parallel to the equipotential surfaces.

A 1 and 3 are correct

B 1 and 2 are correct

C 2 and 4 are correct

D 3 and 4 are correct

💡 Short Explanation

�� Electric field equals potential gradient. �� Potential decreases along the field direction. �� Equipotential surfaces never intersect.

✅ Correct Answer Explanation

The magnitude of the electric field is given by where is the normal displacement. Therefore, statement 1 is correct. Since electric potential decreases in the direction of the electric field, is negative when moving along the field direction, making statement 3 correct. Statement 2 is incorrect because two equipotential surfaces can never intersect; otherwise a point would possess two different potential values simultaneously. Statement 4 is incorrect because electric field lines are always perpendicular, not parallel, to equipotential surfaces.

❓ Why Other Options Are Incorrect

�� Option B → Statement 2 is incorrect.
�� Option C → Statements 2 and 4 are incorrect.
�� Option D → Statement 4 is incorrect.

🎯 Strategy

Used – NCERT Recall

Application
Recall the field-potential relation and properties of equipotential surfaces.
Final Logic
Potential falls along the field and equipotential surfaces never intersect.

🧠 Memory Trick

Field ⟂ Equipotential

📚 Class: XII | Book: NCERT Physics Part I | Chapter: Electrostatic Potential and Capacitance | Topic: Steepest Decrease Principles | Subtopic: Normal Displacement for Potential Magnitude

13 When systematically calculating the electrostatic potential energy of a multi-charge system from scratch, the work done to bring the first charge from infinity is strictly zero because:

A The charge has virtually no resting mass.

B Infinity is treated as relatively close.

C There is no pre-existing external field present to do work against.

D The driving force is purely attractive.

💡 Short Explanation

�� The first charge is placed in empty space. �� No electric field exists initially. �� Therefore no work is needed.

✅ Correct Answer Explanation

During the assembly of a charge system, the first charge is brought from infinity before any other charges are present. Since no electric field exists initially, the charge experiences no electrostatic force during its placement. Therefore, the external agent does not need to perform any work. Work becomes necessary only when subsequent charges are brought into the electric field created by previously placed charges. Hence, the work done in placing the first charge is exactly zero.

❓ Why Other Options Are Incorrect

�� Option A → Rest mass is unrelated to electrostatic assembly work.
�� Option B → Infinity is not treated as close.
�� Option D → No force exists before the first charge is placed.

🎯 Strategy

Used – Logical Analysis

Application
Analyze the system before any charge is introduced.
Final Logic
No field means no force and therefore no work.

🧠 Memory Trick

First Charge → Free Placement

📚 Class: XII | Book: NCERT Physics Part I | Chapter: Electrostatic Potential and Capacitance | Topic: Two Charge System Energy | Subtopic: Zero Work Step for Initial Charge Placement

14 Potential energy mechanics for like and unlike pairs:
Statements:
1. For like pairs, positive work is physically needed against the repulsive force.
2. For unlike pairs, positive work is needed for the reverse path (taking charges to infinity).
3. The formula inherently handles both algebraic signs perfectly.
4. The stored potential energy depends heavily on how rapidly the charges were assembled.

A 1, 2 and 3 are correct

B 2 and 4 are correct

C 1 and 4 are correct

D 3 and 4 are correct

💡 Short Explanation

�� Like charges require positive work to assemble. �� Unlike charges require energy to separate. �� Potential energy is independent of assembly speed.

✅ Correct Answer Explanation

The electrostatic potential energy of two charges is For like charges, , giving positive potential energy. External work must be done against repulsion to bring them closer. For unlike charges, , giving negative potential energy. To separate such charges and move them to infinity, positive work must be supplied. The formula automatically accounts for the correct sign through the product . Statement 4 is incorrect because electrostatic potential energy depends only on the configuration, not on the rate of assembly.

❓ Why Other Options Are Incorrect

�� Option B → Statement 4 is incorrect.
�� Option C → Statement 4 is incorrect.
�� Option D → Statement 4 is incorrect.

🎯 Strategy

Used – Concept Application

Application
Analyze the sign of in the potential energy expression.
Final Logic
The sign of the charge product determines the nature of the stored energy.

🧠 Memory Trick

Unlike → Negative U

📚 Class: XII | Book: NCERT Physics Part I | Chapter: Electrostatic Potential and Capacitance | Topic: Two Charge System Energy | Subtopic: Potential Energy for Like and Unlike Pairs

15 In assembling three isolated charges from infinity, the total stored energy is determined by calculating:

D Zero

💡 Short Explanation

�� Total energy equals total work done. �� Individual work contributions are added. �� Superposition principle applies.

✅ Correct Answer Explanation

The electrostatic potential energy stored in a system of charges equals the total work done in assembling the system from infinity. If , , and represent the work done during successive stages of assembly, then the total stored energy is obtained by summing all contributions: The first charge generally requires zero work, but the total expression still follows the principle of adding all individual work contributions. This additive property arises from the superposition principle and the conservative nature of electrostatic forces.

❓ Why Other Options Are Incorrect

�� Option A → Work contributions are not subtracted.
�� Option B → Energies are added, not multiplied.
�� Option D → Multi-charge systems generally possess non-zero energy.

🎯 Strategy

Used – NCERT Recall

Application
Recall the assembly method used to calculate electrostatic potential energy.
Final Logic
Total energy equals the sum of all assembly work contributions.

🧠 Memory Trick

Total Energy = Total Work Added

📚 Class: XII | Book: NCERT Physics Part I | Chapter: Electrostatic Potential and Capacitance | Topic: Three Charge System Energy | Subtopic: Cumulative Work for Multi-Charge Assembly

16 Three equal charges of are systematically brought from infinity to the corners of an equilateral triangle of side . The total work done is:

💡 Short Explanation

�� Total energy equals sum of pair interaction energies. �� Three charges form three interaction pairs. �� Use electrostatic potential energy formula.

✅ Correct Answer Explanation

For three equal charges placed at the corners of an equilateral triangle, the total electrostatic potential energy is the sum of the potential energies of all unique pairs. Number of pairs: Potential energy of one pair: Since there are three identical pairs, Therefore, the total work done in assembling the configuration is

❓ Why Other Options Are Incorrect

�� Option A → Considers only one interaction pair.
�� Option B → Like charges require positive work for assembly.
�� Option C → Triple counting of pair interactions.

🎯 Strategy

Used – Substitution

Application
Calculate the energy of one pair and multiply by the number of unique pairs.
Final Logic
Three charges create three equal interaction pairs.

🧠 Memory Trick

3 Charges → 3 Pairs

📚 Class: XII | Book: NCERT Physics Part I | Chapter: Electrostatic Potential and Capacitance | Topic: Three Charge System Energy | Subtopic: Interaction Pairs in Potential Energy Formula

17 Incorrect statement concerning energy configurations in external fields:

A The external field E is generally assumed to be a known or specified variable.

B The external field E is directly produced by the charge q whose potential energy we are actively calculating.

C We deliberately ignore the internal potential energy of the external sources themselves.

D Potential V at a given point is the work done bringing a unit positive charge from infinity.

💡 Short Explanation

�� External field is produced by external sources. �� Test charge does not create the external field. �� Potential energy is evaluated within a pre-existing field.

✅ Correct Answer Explanation

When calculating the potential energy of a charge in an external field, the field is assumed to be generated by sources external to the charge under consideration. The charge whose energy is being calculated is treated as a test charge and is assumed not to significantly alter the source configuration. Therefore, statement B is incorrect because the external field is not produced by the charge whose potential energy is being evaluated. Statements A, C and D correctly describe the assumptions and definitions used in external-field energy calculations.

❓ Why Other Options Are Incorrect

�� Option A → Correct assumption in external field problems.
�� Option C → Standard approximation used in electrostatics.
�� Option D → Correct definition of electric potential.

🎯 Strategy

Used – NCERT Recall

Application
Recall the meaning of an external electric field.
Final Logic
External field must originate from sources other than the given charge.

🧠 Memory Trick

External Means Other Sources

📚 Class: XII | Book: NCERT Physics Part I | Chapter: Electrostatic Potential and Capacitance | Topic: Energy in External Electric Fields | Subtopic: Distinguishing External Fields from Source Fields

18 The formal potential energy of a system of two charges and located at and respectively, within an external field , is given by:

💡 Short Explanation

�� External field contributes energy. �� Mutual interaction contributes energy. �� Total energy is the sum of both parts.

✅ Correct Answer Explanation

For two charges placed in an external electric field, the total electrostatic potential energy consists of two contributions: 1. Energy due to the external potential: 1. Mutual interaction energy: Adding both contributions, This expression accounts for both the influence of the external field and the electrostatic interaction between the charges themselves.

❓ Why Other Options Are Incorrect

�� Option A → Omits mutual interaction energy.
�� Option B → Omits energy due to the external field.
�� Option D → Total energy is generally non-zero.

🎯 Strategy

Used – Concept Application

Application
Identify all sources contributing to electrostatic potential energy.
Final Logic
Total energy = External-field energy + Interaction energy.

🧠 Memory Trick

External + Mutual = Total

📚 Class: XII | Book: NCERT Physics Part I | Chapter: Electrostatic Potential and Capacitance | Topic: Energy in External Electric Fields | Subtopic: Potential Energy Calculation for Single Charge

19 Correct statements regarding the Electron Volt unit parameters:
Statements:
1. It intrinsically measures energy, not electric potential.
2. It is mathematically equivalent to .
3. It is used frequently to calculate macroscopic classical mechanical friction.

A 1 and 2 are correct

B 2 and 3 are correct

C 1 and 3 are correct

D 1, 2 and 3 are correct

💡 Short Explanation

�� Electron volt is a unit of energy. �� It has a fixed joule equivalent. �� It is mainly used in atomic and nuclear physics.

✅ Correct Answer Explanation

An electron volt (eV) is defined as the energy gained by an electron when accelerated through a potential difference of 1 volt. Therefore, it is fundamentally a unit of energy rather than electric potential. Thus statements 1 and 2 are correct. Statement 3 is incorrect because electron volts are generally used in atomic, nuclear and particle physics where energies are extremely small. Macroscopic friction problems are typically analyzed using joules rather than electron volts.

❓ Why Other Options Are Incorrect

�� Option B → Statement 3 is incorrect.
�� Option C → Statement 3 is incorrect.
�� Option D → Statement 3 is incorrect.

🎯 Strategy

Used – NCERT Recall

Application
Recall the definition and usage of the electron volt.
Final Logic
Electron volt is an energy unit used in microscopic physics.

🧠 Memory Trick

eV = Electron Energy Value

📚 Class: XII | Book: NCERT Physics Part I | Chapter: Electrostatic Potential and Capacitance | Topic: Electron Volt Energy Unit | Subtopic: Definition Based on Elementary Charge and 1 Volt

20 Primary usage domain of the electron volt (eV) energy unit:

A Macroscopic gravitational physics

B Atomic, nuclear, and particle physics

C Bulk fluid thermodynamics

D Simple projectile kinematics

💡 Short Explanation

�� Electron volt is a microscopic energy unit. �� Widely used for atoms and nuclei. �� Convenient for particle interactions.

✅ Correct Answer Explanation

The electron volt is an energy unit specifically suited for describing very small energies encountered in microscopic systems. Atomic transitions, nuclear reactions and particle interactions typically involve energies ranging from electron volts to giga-electron volts. Using joules for such quantities would lead to extremely small numbers. Therefore, physicists commonly use eV, keV, MeV and GeV in atomic, nuclear and particle physics. This makes calculations and comparisons much more convenient.

❓ Why Other Options Are Incorrect

�� Option A → Gravitational systems generally use joules.
�� Option C → Thermodynamics commonly uses joules and calories.
�� Option D → Projectile motion problems are usually expressed in SI units.

🎯 Strategy

Used – NCERT Recall

Application
Recall the standard applications of the electron volt.
Final Logic
Electron volt is designed for microscopic energy scales.

🧠 Memory Trick

eV → Electrons, Atoms, Nuclei

📚 Class: XII | Book: NCERT Physics Part I | Chapter: Electrostatic Potential and Capacitance | Topic: Electron Volt Energy Unit | Subtopic: Conversion to Joules and Atomic Physics Usage

CUET UG Physics Booster Test – 3 Equipotentials and Energy

CUET UG Physics Booster Test - 3 Equipotentials and Energy

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