CUET UG Physics Booster Test - 2 Equipotentials and Energy -

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CUET UG Physics Booster Test - 2 Equipotentials and Energy

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Physics Booster Test - 2 Equipotentials and Energy

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QUESTION 1 OF 20

Incorrect statement about defining equipotential surfaces:

Answer Review

1 Incorrect statement about defining equipotential surfaces:

A The surface map varies depending on the specific charge configuration.

B Potential V is constant if distance r is constant for a point charge.

C Work is required to move a charge between two points on the same surface.

D There is no potential difference between any two points on the same surface.

💡 Short Explanation

�� Equipotential surfaces have constant potential. �� Potential difference along the surface is zero. �� Therefore, no work is required.

✅ Correct Answer Explanation

An equipotential surface is a surface on which the electric potential remains constant at every point. Since potential difference between any two points on such a surface is zero, the work done in moving a charge along the surface is also zero. Since , Therefore, statement C is incorrect. Statement A is correct because the shape of equipotential surfaces depends on the charge distribution producing the field. Statement B is correct for a point charge because all points at the same radial distance have the same potential. Statement D is also correct because an equipotential surface has no potential difference between any two of its points.

❓ Why Other Options Are Incorrect

�� Option A → Correct property of equipotential maps.
�� Option B → Correct for a point charge.
�� Option D → Correct definition of an equipotential surface.

🎯 Strategy

Used – Concept Application

Application
Apply the definition of equipotential surfaces and the work-potential relation.
Final Logic
No potential difference means no work done.

🧠 Memory Trick

Equipotential → Equal Potential → Zero Work

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

2 To move a test charge q between any two points on an equipotential surface, the total required work W is strictly:

A qV

B −qV

C Infinite

D 0

💡 Short Explanation

�� Equipotential means constant potential. �� Potential difference is zero. �� Therefore work done is zero.

✅ Correct Answer Explanation

The work done in moving a charge between two points is On an equipotential surface, all points have the same electric potential. Therefore, Substituting into the work formula, This means that moving a charge anywhere along an equipotential surface requires no work. This property is one of the defining characteristics of equipotential surfaces and is widely used in electrostatic analysis.

❓ Why Other Options Are Incorrect

�� Option A → Potential difference, not potential itself, determines work.
�� Option B → Incorrect expression for work.
�� Option C → No infinite work is involved.

🎯 Strategy

Used – Substitution

Application
Substitute ΔV = 0 into the work formula.
Final Logic
Zero potential difference gives zero work.

🧠 Memory Trick

Same Potential → Same Energy → No 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 Match the following:

List I	List II
1. Single positive point charge	a. Radial inward field lines, equipotentials are concentric spheres
2. Single negative point charge	b. Radial outward field lines, equipotentials are concentric spheres
3. Positive charge field direction	c. Away from the charge
4. Negative charge field direction	d. Toward the charge B.1-b, 2-a, 3-c, 4-d

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

B (option not parsed — please edit)

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

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

💡 Short Explanation

�� Equipotential surfaces are spherical for point charges. �� Positive charges produce outward fields. �� Negative charges produce inward fields.

✅ Correct Answer Explanation

For an isolated point charge, the electric potential depends only on radial distance from the charge. Therefore, equipotential surfaces are concentric spheres centered on the charge. For a positive charge, electric field lines emerge outward from the charge. For a negative charge, electric field lines terminate inward toward the charge. Thus the correct matching associates positive charges with outward radial fields and negative charges with inward radial fields while maintaining spherical equipotential surfaces.

❓ Why Other Options Are Incorrect

�� Option A → Positive and negative charge properties are interchanged.
�� Option C → Incorrect assignment of field directions.
�� Option D → Multiple mismatches between charge type and field orientation.

🎯 Strategy

Used – NCERT Recall

Application
Recall the field-line pattern of isolated point charges.
Final Logic
Positive charges push outward; negative charges pull inward.

🧠 Memory Trick

Negative → Pulls In

📚 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 Parallel planes for uniform electric fields:
1. Are planes normal to the uniform field direction.
2. Are planes parallel to the uniform field direction.
3. Have constant potential on each individual plane.
4. Intersect each other.

A 1 and 3 are correct

B 2 and 3 are correct

C 1 and 4 are correct

D 3 and 4 are correct

💡 Short Explanation

�� Equipotential surfaces are perpendicular to field lines. �� Uniform fields produce parallel equipotential planes. �� Each plane has constant potential.

✅ Correct Answer Explanation

In a uniform electric field, the electric field lines are parallel and equally spaced. Equipotential surfaces must always be perpendicular to the electric field. Therefore, equipotential surfaces are planes normal to the field direction. Every point on a given plane possesses the same electric potential, making it an equipotential surface. Different equipotential planes never intersect because a point cannot simultaneously have two different potentials. Hence statements 1 and 3 are correct, while statements 2 and 4 are incorrect.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used – Concept Application

Application
Use the relationship between electric field lines and equipotential surfaces.
Final Logic
Equipotential planes are perpendicular to the field and do not intersect.

🧠 Memory Trick

Field ⟂ Equipotential

📚 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 Normal orientation setup statements:
1. The component of the electric field along the equipotential surface must be zero.
2. This normal orientation follows directly from the zero potential difference property.
3. This principle applies only to isolated point charges, not dipoles.
4. If the field was not normal, it would have some non-zero component along the surface.

A 3 and 4 are correct

B 2 and 4 are correct

C 1, 2 and 3 are correct

D 1, 2 and 4 are correct

💡 Short Explanation

�� Electric field is normal to equipotential surfaces. �� Tangential field components must vanish. �� The rule applies to all electrostatic fields.

✅ Correct Answer Explanation

Equipotential surfaces have the same potential at every point. Therefore, moving a charge along the surface involves no change in potential and no work is done. If an electric field had a component along the surface, work would be done on a moving charge, producing a potential difference. This contradicts the definition of an equipotential surface. Hence the tangential component of the electric field must be zero, making the field normal to the surface. This principle is universal and applies to point charges, dipoles and all electrostatic configurations. Therefore statements 1, 2 and 4 are correct, while statement 3 is incorrect.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used – Logical Analysis

Application
Examine what would happen if the field had a tangential component.
Final Logic
A tangential field would create a potential difference, which is impossible on an equipotential surface.

🧠 Memory Trick

Equipotential Surface → No Tangential Field

📚 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 Consequence of the non-existence of tangential field components:

A Zero work to move a charge on the surface

B Infinite potential difference on the surface

C Electric field lines cross each other

D Charges continuously accelerate along the surface

💡 Short Explanation

�� Tangential electric field is zero. �� Potential remains constant on the surface. �� No work is done along the surface.

✅ Correct Answer Explanation

An equipotential surface is characterized by a constant electric potential at every point. Since the electric field has no tangential component along the surface, a charge moving on the surface experiences no force in the direction of motion. Consequently, the work done in moving a charge along the equipotential surface is zero. This follows directly from Since for any two points on the surface, The absence of tangential electric field components is therefore responsible for the zero-work property of equipotential surfaces.

❓ Why Other Options Are Incorrect

�� Option B → Potential difference is zero, not infinite.
�� Option C → Electric field lines never cross each other.
�� Option D → No tangential force exists to cause continuous acceleration.

🎯 Strategy

Used – Concept Application

Application
Relate the absence of tangential electric field to the work-potential relationship.
Final Logic
No tangential field means no work along the surface.

🧠 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: Non-existence of Tangential Field Components

7 Correct statements about the surface map for electric dipoles:
1. The equatorial plane acts as a zero-potential surface.
2. The map is perfectly symmetrical along the equatorial plane.
3. The equipotential surfaces are perfectly concentric spheres everywhere.

A 1 and 3 are correct

B 1 and 2 are correct

C 2 and 3 are correct

D 3 only is correct

💡 Short Explanation

�� Dipole potential is zero on the equatorial plane. �� Equipotential patterns show symmetry. �� Dipole surfaces are not spherical.

✅ Correct Answer Explanation

An electric dipole consists of equal and opposite charges separated by a small distance. Due to symmetry, the equatorial plane is a zero-potential surface because the positive and negative charge contributions cancel exactly. The equipotential map is symmetric about the equatorial plane. However, unlike the case of a single point charge, the equipotential surfaces are not concentric spheres. Their shapes are distorted due to the presence of two opposite charges. Therefore statements 1 and 2 are correct, while statement 3 is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Statement 3 is incorrect.
�� Option C → Statement 3 is incorrect.
�� Option D → Statements 1 and 2 are also correct.

🎯 Strategy

Used – NCERT Recall

Application
Recall the standard equipotential map of an electric dipole.
Final Logic
Dipole equipotential surfaces are symmetric but not spherical.

🧠 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 For two identical positive charges, the equipotential surfaces very close to each individual charge are:

A Nearly spherical

B Perfectly flat planes

C Perfectly straight lines

D Large overarching cylinders

💡 Short Explanation

�� Nearby regions are dominated by one charge. �� Potential resembles that of a point charge. �� Equipotential surfaces become nearly spherical.

✅ Correct Answer Explanation

When observing points very close to one of two identical positive charges, the contribution of the nearby charge is much greater than that of the distant charge. Therefore, the electric potential in that region behaves almost like the potential due to a single isolated point charge. Since equipotential surfaces around a point charge are concentric spheres, the surfaces near each charge are approximately spherical. At larger distances, the influence of both charges becomes significant and the equipotential surfaces become more complex.

❓ Why Other Options Are Incorrect

�� Option B → Equipotential surfaces are not flat near a point charge.
�� Option C → Straight lines cannot represent three-dimensional equipotential surfaces.
�� Option D → Cylindrical surfaces do not describe the local geometry around a point charge.

🎯 Strategy

Used – Concept Application

Application
Identify which charge dominates the potential in the nearby region.
Final Logic
Near one charge, the field resembles that of a single point charge.

🧠 Memory Trick

Near Charge → Near Sphere

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

9 The mathematical relation between field and potential concludes that the electric field points:

A Parallel to the constant potential surfaces

B In the direction in which the potential decreases steepest

C In the direction in which potential increases steepest

D Independent of the potential gradient

💡 Short Explanation

�� Electric field equals negative potential gradient. �� Field points toward lower potential. �� Maximum decrease determines direction.

✅ Correct Answer Explanation

The relation between electric field and electric potential is The negative sign indicates that the electric field points toward decreasing potential. More specifically, it points in the direction where the potential decreases most rapidly. This relationship explains why electric field lines are perpendicular to equipotential surfaces. The field always seeks the steepest fall in potential and therefore determines the direction of motion of a positive test charge.

❓ Why Other Options Are Incorrect

�� Option A → Electric field is normal to equipotential surfaces.
�� Option C → Field points toward decreasing potential.
�� Option D → Electric field is directly related to the potential gradient.

🎯 Strategy

Used – NCERT Recall

Application
Recall the gradient relationship between field and potential.
Final Logic
The negative sign in determines the direction.

🧠 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 If the potential decreases by 20 V over a normal displacement of 2 mm, the magnitude of the uniform electric field is:

A 10⁴ V/m

B 10 V/m

C 40 V/m

D 40 × 10⁻³ V/m

💡 Short Explanation

�� Use . �� Convert mm into metres. �� Calculate the potential gradient.

✅ Correct Answer Explanation

The magnitude of the electric field is given by Given: Substituting, Therefore, the magnitude of the electric field is This result follows directly from the definition of electric field as the rate of decrease of potential with distance.

❓ Why Other Options Are Incorrect

�� Option B → Unit conversion error.
�� Option C → Incorrect division.
�� Option D → Wrong numerical magnitude.

🎯 Strategy

Used – Substitution

Application
Apply the formula after converting units.
Final Logic
Divide the potential change by the normal displacement.

🧠 Memory Trick

Field = Voltage Drop ÷ 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 The equation
mathematically proves that:

A Field lines must act parallel to equipotentials

B Potential is a vector quantity

C Field magnitude equals the change in potential magnitude per unit normal displacement

D Field is always a scalar value

💡 Short Explanation

�� Electric field is related to potential gradient. �� Potential changes most rapidly along the normal direction. �� Field magnitude equals rate of potential decrease.

✅ Correct Answer Explanation

The relation connects electric field magnitude with the rate of change of electric potential. Here, represents displacement in the direction of the electric field, which is normal to an equipotential surface. The negative sign indicates that electric potential decreases in the direction of the electric field. Therefore, the electric field magnitude is numerically equal to the decrease in potential per unit normal displacement. This equation forms the mathematical link between electric field and electric potential and is extensively used in electrostatics.

❓ Why Other Options Are Incorrect

�� Option A → Electric field is perpendicular, not parallel, to equipotential surfaces.
�� Option B → Electric potential is a scalar quantity.
�� Option D → Electric field is a vector quantity.

🎯 Strategy

Used – Concept Application

Application
Interpret each term in the equation relating field and potential.
Final Logic
Electric field magnitude equals potential drop per unit normal distance.

🧠 Memory Trick

Field = Potential Drop ÷ Distance

📚 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 for potential magnitude statements:
1. calculates the field component.
2. is inherently negative in the direction of the electric field.
3. The field opposes normal displacement completely.

A 1 and 2 are correct

B 2 and 3 are correct

C 3 and 4 are correct

D 1 and 4 are correct

💡 Short Explanation

�� Potential decreases along the field direction. �� Electric field equals negative potential gradient. �� Correct mathematical relation is important.

✅ Correct Answer Explanation

The electric field magnitude is related to potential through This equation states that electric field is the negative rate of change of potential with distance. Since potential decreases in the direction of the electric field, is negative when displacement is taken along the field direction. Therefore statements 1 and 2 are correct. Statement 3 is incorrect because it reverses the correct relation. Statement 4 is also incorrect because the electric field does not necessarily oppose normal displacement; its effect depends on the direction of motion and the sign of the charge.

❓ Why Other Options Are Incorrect

�� Option B → Statement 3 is incorrect.
�� Option C → Statements 3 and 4 are incorrect.
�� Option D → Statement 4 is incorrect.

🎯 Strategy

Used – NCERT Recall

Application
Recall the standard field-potential relationship.
Final Logic
Potential decreases in the direction of the electric field.

🧠 Memory Trick

Minus Sign → Potential Falls

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

13 When building a two-charge system, why is the external work done in bringing the first charge from infinity to its location exactly zero?

A Because infinity mathematically possesses infinite potential

B Because there is no external field present against which work needs to be done

C Because the magnitude of a single charge is considered negligible

D Because the electrostatic force is non-conservative initially

💡 Short Explanation

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

✅ Correct Answer Explanation

When assembling a system of charges, the first charge is brought from infinity before any other charges are present. Since no electric field exists at that stage, there is no electrostatic force opposing or assisting the motion of the charge. Consequently, the external agent does not need to perform any work to place the first charge at its designated location. The work done becomes important only when subsequent charges are brought into the electric field produced by previously placed charges.

❓ Why Other Options Are Incorrect

�� Option A → Potential at infinity is taken as zero, not infinite.
�� Option C → The charge magnitude is irrelevant to this argument.
�� Option D → Electrostatic force is conservative throughout.

🎯 Strategy

Used – Logical Analysis

Application
Consider the state of the system before any charge is placed.
Final Logic
No field means no force and therefore no work.

🧠 Memory Trick

First Charge → No Opposition

📚 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 Incorrect statement about two-charge potential energy:

A If q₁q₂ > 0, the electrostatic potential energy is positive.

B If q₁q₂ < 0, the electrostatic potential energy is negative.

C The final potential energy expression depends fundamentally on the specific path taken to bring the charges together.

D The work done gets fully stored in the form of potential energy.

💡 Short Explanation

�� Electrostatic force is conservative. �� Potential energy depends only on initial and final states. �� Path independence is a key property.

✅ Correct Answer Explanation

The electrostatic potential energy of two charges separated by distance r is This expression depends only on the charges and their separation. Since electrostatic force is conservative, the work done in assembling the system is independent of the path followed. Therefore statement C is incorrect. Statements A and B correctly describe the sign of potential energy for like and unlike charges respectively. Statement D is also correct because the work done by the external agent is stored as electrostatic potential energy.

❓ Why Other Options Are Incorrect

�� Option A → Correct sign convention for like charges.
�� Option B → Correct sign convention for unlike charges.
�� Option D → Correct interpretation of stored energy.

🎯 Strategy

Used – Concept Application

Application
Use the conservative nature of electrostatic forces.
Final Logic
Potential energy depends only on configuration, not on path.

🧠 Memory Trick

Conservative Force → Path Independent

📚 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 Cumulative work for assembling three charges q₁, q₂, q₃:
1. Bringing q₃ requires work against the combined fields of q₁ and q₂.
2. The initial work to place q₁ is strictly non-zero.
3. The total cumulative work is completely independent of the assembly order.
4. The final energy relies solely on external driving forces.

A 1 and 3 are correct

B 1, 2 and 3 are correct

C 2 and 4 are correct

D 3 and 4 are correct

💡 Short Explanation

�� Third charge experiences the fields of existing charges. �� First charge requires zero work. �� Final energy is path independent.

✅ Correct Answer Explanation

While assembling a three-charge system, the first charge is placed without doing any work because no electric field exists initially. Therefore statement 2 is incorrect. When bringing the third charge q₃, it experiences the combined electric fields produced by q₁ and q₂, making statement 1 correct. Since electrostatic forces are conservative, the total work required to assemble the system depends only on the final configuration and not on the order of assembly. Hence statement 3 is also correct. Statement 4 is incorrect because the final electrostatic energy is determined by charge interactions rather than external driving forces alone.

❓ 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 – Concept Application

Application
Analyze the assembly process step by step.
Final Logic
Only previously placed charges contribute to the work required for later charges.

🧠 Memory Trick

First Free, Third Feels Two

📚 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 In the potential energy formula for three charges, the final expression is independent of the manner in which the configuration is assembled because of:

A The friction limits

B The non-conservative nature of assembly

C The conservative nature of the electrostatic force

D The universally zero potential at the origin

💡 Short Explanation

�� Electrostatic force is conservative. �� Work depends only on initial and final states. �� Assembly order does not affect total energy.

✅ Correct Answer Explanation

The electrostatic force is a conservative force. Therefore, the work done in moving charges depends only on their initial and final positions and not on the path followed. While assembling a system of three charges, different assembly sequences may involve different intermediate steps, but the total work done remains the same. Consequently, the electrostatic potential energy of the final configuration is independent of the manner in which the charges are assembled. This property allows a unique expression for the potential energy of the system and follows directly from the conservative nature of electrostatic interactions.

❓ Why Other Options Are Incorrect

�� Option A → Friction is not involved in electrostatic potential energy calculations.
�� Option B → Electrostatic force is conservative, not non-conservative.
�� Option D → The origin does not determine the assembly energy.

🎯 Strategy

Used – NCERT Recall

Application
Recall the characteristics of conservative forces.
Final Logic
Conservative forces make total work path independent.

🧠 Memory Trick

Conservative Force → Same Final Energy

📚 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 In determining the potential energy of a charge in an external field, what is critically assumed about the test charge q?

A It significantly alters the external sources generating the field.

B It alone produces the entire external field.

C It does not significantly affect the sources producing the external field.

D It rapidly neutralizes the local field entirely.

💡 Short Explanation

�� The charge is treated as a test charge. �� External sources remain unchanged. �� Potential energy is calculated using the existing field.

✅ Correct Answer Explanation

When calculating the potential energy of a charge in an external electric field, the charge is assumed to be small enough that it does not significantly disturb the sources producing the field. This allows the external field to remain unchanged during the calculation. Under this assumption, if the external potential at a point is V and the charge is q, the potential energy is This approximation is fundamental in electrostatics and permits the independent analysis of a charge placed in a pre-existing electric field.

❓ Why Other Options Are Incorrect

�� Option A → A test charge is assumed not to alter the external field significantly.
�� Option B → The external field is produced by external sources.
�� Option D → A test charge does not neutralize the field.

🎯 Strategy

Used – NCERT Recall

Application
Recall the assumptions involved in defining a test charge.
Final Logic
The external field must remain essentially unchanged.

🧠 Memory Trick

Test Charge → Tiny Influence

📚 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 If an electron is placed at a position where the external potential is 10 V, its potential energy is:

A 1.6 × 10⁻¹⁸ J

B −1.6 × 10⁻¹⁸ J

C 1.6 × 10⁻²⁰ J

D −1.6 × 10⁻¹⁹ J

💡 Short Explanation

�� Use U = qV. �� Electron charge is negative. �� Potential energy becomes negative.

✅ Correct Answer Explanation

Potential energy of a charge in an external electric field is For an electron, Given, Substituting, The negative sign arises because the electron carries negative charge. Therefore, the potential energy of the electron at that point is

❓ Why Other Options Are Incorrect

�� Option A → Sign of electron charge is ignored.
�� Option C → Incorrect power of ten.
�� Option D → Calculation error in multiplication.

🎯 Strategy

Used – Substitution

Application
Apply the formula U = qV using the electron charge.
Final Logic
Negative charge produces negative potential energy for positive potential.

🧠 Memory Trick

Electron = Negative Charge = Negative U

📚 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 about the Electron Volt:
1. It is fundamentally a unit of power, not energy.
2. It is defined as the energy gained by an electron accelerated by a 1 volt potential difference.
3. 1 eV = 1.6 × 10⁻¹⁹ J.

A 2 and 3 are correct

B 1 and 2 are correct

C 1 and 3 are correct

D 3 only is correct

💡 Short Explanation

�� Electron volt is a unit of energy. �� Defined using an electron and 1 volt. �� Widely used in atomic and nuclear physics.

✅ Correct Answer Explanation

An electron volt (eV) is defined as the energy gained by an electron when it moves through a potential difference of 1 volt. Since the charge of an electron is the corresponding energy is Therefore, statements 2 and 3 are correct. Statement 1 is incorrect because electron volt is a unit of energy, not power. It is frequently used in atomic, nuclear and particle physics because the joule is too large for microscopic energy scales.

❓ Why Other Options Are Incorrect

�� Option B → Statement 1 is incorrect.
�� Option C → Statement 1 is incorrect.
�� Option D → Statement 2 is also correct.

🎯 Strategy

Used – NCERT Recall

Application
Recall the definition and SI conversion of the electron volt.
Final Logic
Electron volt measures energy, not power.

🧠 Memory Trick

1 Volt + Electron = 1 eV

📚 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 Conversion of 1 GeV to Joules:

A 1.6 × 10⁻¹⁹ J

B 1.6 × 10⁻¹³ J

C 1.6 × 10⁻¹⁶ J

D 1.6 × 10⁻¹⁰ J

💡 Short Explanation

�� 1 eV = 1.6 × 10⁻¹⁹ J. �� 1 GeV = 10⁹ eV. �� Multiply the conversion factor accordingly.

✅ Correct Answer Explanation

The standard conversion is Since Therefore, GeV is a commonly used energy unit in particle physics because elementary particles often possess energies in the giga-electron-volt range.

❓ Why Other Options Are Incorrect

�� Option A → Corresponds to 1 eV.
�� Option B → Corresponds to 1 MeV.
�� Option C → Incorrect power of ten.

🎯 Strategy

Used – Substitution

Application
Convert GeV into eV and then into joules.
Final Logic
.

🧠 Memory Trick

GeV → Add 9 to the Exponent

📚 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 – 2 Equipotentials and Energy

CUET UG Physics Booster Test - 2 Equipotentials and Energy

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