Chpt 2 Gussacs And Ekectric Potential

Question 1

10. Which of the following is correct?A. Equipotential surface is a surface where all points have equal potentialB. The surfaces have the same chargeC. There are no lines of forceD. Point of equal electric intensity

Answer 1

Answer: AExplanation: An equipotential surface is defined as a surface where the electric potential is the same at every point which means that no work is required to move a charge along this surface

Question 2

12. What do you understand by saying that electric charge is quantized

Answer 2

Answer: This is a conceptual questioExplanation: Electric charge is quantized means that charge exists in discrete packets rather than a continuous quantity. This is due to the fact that charge is carried by particles like electrons and protons which have fixed indivisible charges

Question 3

13. Define the dielectric constant of a medium

Answer 3

Answer: This is a conceptual questionExplanation: The dielectric constant of a medium is the ratio of the permittivity of the medium to the permittivity of free space. It is a measure of a material’s ability to store electrical energy in an electric field

Question 4

15. Compute the ratio of electric to gravitational force between two protons

Answer 4

Answer: This is a conceptual question involving calculationExplanation: The electric force between two protons can be calculated using Coulomb’s law and the gravitational force can be calculated using Newton’s law of gravitation. The ratio of these forces shows that electric force is vastly stronger than gravitational force at the particle level

Question 5

16. Point charges and are placed in a straight line. The central one is from each of the others. Calculate the net force on each due to the other two

Answer 5

Answer: Calculation requiredExplanation: Use Coulomb’s law to calculate the force between each pair of charges and then use vector addition to find the net force on each charge

Question 6

17. Two charges and are at distance apart. These two charges are free to move but do not because there is a third charge nearby. What must be the charge and placement of the third charge for the first two to be in equilibrium?

Answer 6

Answer: This is a conceptual problem involving equilibrium conditionsExplanation: The third charge must create a force that counterbalances the repulsive force between the two like charges. Its magnitude and position depend on solving the force balance equations

Question 7

1. Which of the following statements is not correct?A. There are two types of electric charge positive and negative chargeB. Materials such as polythene and cellulose acetate can be charged by frictionC. The positive charge on the woolen material is not of the same magnitude as the charge on the polytheneD. Electric charge is always conserved in a closed system

Answer 7

Answer: CExplanation: The magnitude of charge created by friction on two objects should be equal and opposite due to the conservation of charge. If one object gains a positive charge the other gains an equal negative charge

Question 8

2. Which of the following is not correctA. Water is an insulatorB. Carbon is an insulatorC. Plastic is a semi-conductorD. Copper is a good conductor

Answer 8

Answer: CExplanation: Plastic is generally considered an insulator not a semiconductor. Semiconductors are materials like silicon or germanium which have properties between conductors and insulators

Question 9

3.which is incorrect Dielectric constant of:A. Vacuum is B. Air is close to 1C. Metal is infinityD. Water is 0

Answer 9

Answer: DExplanation: The dielectric constant of water is actually much higher than 1 typically around 80 at room temperature so “Water is 0” is incorrect

Question 10

4. Which of the following statements is correct regarding water?A. Good conductorB. Poor conductorC. Semi-conductorD. Insulator

Answer 10

Answer: BExplanation: Pure water is a poor conductor of electricity but it can conduct electricity when impurities or ions are present

Question 11

5. Silver is a:A. Good conductorB. Poor conductorC. Semi-conductorD. Insulator

Answer 11

Answer: AExplanation: Silver is one of the best conductors of electricity due to its high conductivity and low resistivity

Question 12

6. Carbon is a:A. Good conductorB. Poor conductorC. Semi-conductorD. Insulator

Answer 12

Answer: CExplanation: Carbon particularly in its graphite form can act as a semi-conductor though in diamond form it is an insulator

Question 13

7. Charging by induction is:A. Touching two positive chargesB. Touching two negative chargesC. Charging a body without touching itD. Touching two iron rods

Answer 13

Answer: CExplanation: Induction is a method of charging a conductor by bringing it near another charged body without direct contact causing a separation of charges

Question 14

9. Which of the following is incorrect?A. Electric field intensityB. The unit of electric field intensity isC. The unit of electric field intensity isD. Electric field intensity is a scalar quantity

Answer 14

Answer: DExplanation: Electric field intensity is a vector quantity not a scalar because it has both magnitude and direction

Question 15

10. Which of the following is correct?A. Equipotential surface is a surface where all points have equal potentialB. The surfaces have the same chargeC. There are no lines of forceD. Point of equal electric intensity

Answer 15

Answer: AExplanation: An equipotential surface is defined as a surface where the electric potential is the same at every point which means that no work is required to move a charge along this surface

Question 16

19. A quadrupole consists of two dipoles that are close to each other as shown in Figure 1.18. The effective charge at the origin is and the other charges on the y-axis at and are each(i) Find the electric field at a point on the x-axis far away so that(ii) Find the electric field on the y-axis far so that

Answer 16

Answer: This involves calculating the electric field due to a quadrupole arrangementExplanation: At points far from a quadrupole the electric field decreases faster than for a dipole. Use the quadrupole field equations to approximate the electric field at distant points on both the x-axis and y-axis

Question 17

1. What is the unit vector always perpendicular to the surface area element dS(a) n(b) E(c) Φ(d) λ

Answer 17

Answer (a) nExplanation The unit vector n is defined as always being perpendicular to the surface area element dS

Question 18

2. If the electric field is parallel to the surface area what is the electric flux(a) Maximum(b) Minimum(c) Zero(d) Undefined

Answer 18

Answer (c) ZeroExplanation When the electric field is parallel to the surface the angle between E and n is 90° and cos(90°) = 0 Therefore the flux is zero

Question 19

3. The electric flux through a surface is proportional to what(a) The area of the surface only(b) The electric field strength only(c) The cosine of the angle between E and the surface normal(d) The product of the electric field strength the area and the cosine of the angle between E and the surface normal

Answer 19

Answer (d)Explanation Electric flux depends on all three factors The cosine term accounts for the orientation of the field relative to the surface

Question 20

4. What does the symbol Φ represent in the context of electric flux(a) Electric field(b) Electric flux(c) Surface area(d) Charge density

Answer 20

Answer (b) Electric fluxExplanation Φ denotes electric flux which measures the electric field passing through a given area

Question 21

5. In equation Φ = ∫E ⋅ dS what does the dot product E⋅dS signify(a) The scalar product of E and dS(b) The vector product of E and dS(c) The magnitude of E(d) The magnitude of dS

Answer 21

Answer (a)Explanation The dot product represents the component of the electric field perpendicular to the surface area element which contributes to the flux

Question 22

6. For a closed surface what does Gauss’s law relate(a) Electric flux to enclosed charge(b) Electric potential to charge density(c) Electric field to electric potential(d) Electric field to charge density

Answer 22

Answer (a) Electric flux to enclosed chargeExplanation Gauss’s Law states that the total electric flux through any closed surface is proportional to the net electric charge enclosed within the surface

Question 23

7. What type of charge distribution is equation Φ = ∫E ⋅ dS suited for(a) Discrete charge distribution(b) Continuous charge distribution(c) Both discrete and continuous(d) Neither discrete nor continuous

Answer 23

Answer (b) Continuous charge distributionExplanation The integral form of the equation is best suited for continuous distributions where charge is spread across an area or volume

Question 24

8. In equation E = Q/4πε₀r² what does Q represent(a) Electric field(b) Electric flux(c) Charge enclosed(d) Permittivity of free space

Answer 24

Answer (c) Charge enclosedExplanation Q represents the total charge enclosed within the Gaussian surface

Question 25

9. What is the shape of the Gaussian surface used to derive the electric field of a charged sphere(a) Cylindrical(b) Rectangular(c) Spherical(d) Cubical

Answer 25

Answer (c) SphericalExplanation A spherical Gaussian surface is chosen due to the spherical symmetry of the electric field

Question 26

10. What is the electric field inside a uniformly charged conducting sphere(a) Zero(b) Constant but non-zero(c) Varies with distance from the center(d) Undefined

Answer 26

Answer (a) ZeroExplanation Inside a uniformly charged conducting sphere the electric field is always zero because charges distribute on the surface

Question 27

11. What is the shape of the Gaussian surface used to derive the electric field of an infinitely long charged cylinder(a) Spherical(b) Cubical(c) Cylindrical(d) Rectangular

Answer 27

Answer (c) CylindricalExplanation A cylindrical Gaussian surface is chosen due to the cylindrical symmetry of the field

Question 28

12. For an infinite sheet of charge what is the electric field(a) Zero(b) Constant(c) Depends on distance from the sheet(d) Varies inversely with distance squared

Answer 28

Answer (b) ConstantExplanation The electric field due to an infinite sheet of charge is uniform and does not depend on distance

Question 29

13. What does λ represent in the context of an infinitely long charged cylinder(a) Surface charge density(b) Volume charge density(c) Linear charge density(d) Point charge

Answer 29

Answer (c) Linear charge densityExplanation λ represents the charge per unit length along the cylinder

Question 30

14. Which equation is useful for calculating the electric field when the charge distribution is symmetric(a) Coulomb’s Law(b) Gauss’s Law(c) Ohm’s Law(d) Faraday’s Law

Answer 30

Answer (b) Gauss’s LawExplanation Gauss’s Law is especially useful for symmetric charge distributions

Question 31

15. What is the primary advantage of Gauss’s Law over Coulomb’s Law(a) Gauss’s law works for all charge distributions while Coulomb’s Law doesn’t(b) Gauss’s law is easier to apply to symmetric charge distributions(c) Gauss’s Law provides a more accurate result(d) Coulomb’s Law only works for point charges

Answer 31

Answer (b)Explanation Gauss’s Law simplifies calculations for symmetric charge distributions

Question 32

16. In the example calculation what is the magnitude of the electric field(a) 67.2 Nm²/C(b) 350 NC⁻¹(c) 4πε₀(d) 0

Answer 32

Answer (b) 350 NC⁻¹Explanation The magnitude of the electric field is provided as 350 NC⁻¹ in the problem

Question 33

17. What is the angle between the electric field and the rectangular surface in the example(a) 0°(b) 45°(c) 90°(d) 50°

Answer 33

Answer (d) 50°Explanation The angle between the electric field and the rectangular surface is stated as 50°

Question 34

1. Explain the concept of electric flux and its significance in electrostatics.

Answer 34

Electric flux refers to the measure of the electric field passing through a given area. It is calculated as the product of the electric field and the area it penetrates factoring in the angle between the electric field direction and the normal to the surface. In electrostatics electric flux helps in understanding the behavior of electric fields around charges and is essential in deriving Gauss’s law which relates the flux to the enclosed charge.

Question 35

2. Describe how Gauss’s law simplifies the calculation of electric fields for symmetric charge distributions.

Answer 35

Gauss’s law states that the electric flux through a closed surface is proportional to the enclosed charge. It simplifies electric field calculations for symmetric charge distributions (e.g. spherical cylindrical or planar) by exploiting the symmetry. This law allows us to calculate the electric field without needing to consider every individual charge by using the symmetry to reduce the complexity of the calculations often transforming them into simpler forms using integrals.

Question 36

3. Compare and contrast the use of Gauss’s Law and Coulomb’s Law in calculating electric fields.

Answer 36

Coulomb’s Law is used to calculate the electric field due to a point charge providing a straightforward formula for the force between two point charges. However for complex charge distributions Coulomb’s law becomes cumbersome as it requires summing the contributions from all charges.Gauss’s Law on the other hand is more efficient for charge distributions with symmetry (spherical cylindrical or planar). By selecting an appropriate Gaussian surfacethe law simplifies the calculation of the electric field especially when the charge distribution exhibits symmetry. Gauss’s law is generally more powerful than Coulomb’s law for such cases as it can quickly determine the electric field without dealing with individual charge interactions.

Question 37

2.1 Which of the following statements is incorrect?• A. If a resultant electric field exists in a conductor…• B. Electrons will be accelerated• C. Protons will flow• D. Current will flow

Answer 37

Answer: C. Protons will flowExplanation: In a conductor only electrons are free to move. Protons are bound within the atomic nuclei and cannot contribute to the current flow. If a resultant electric field exists within a conductor the free electrons will accelerate (B) leading to a current flow (D).

Question 38

2.2 Prove that the electric field within a conductor is … A.Q/Eo B.zeroC.maximumD.minimum value

Answer 38

Answer: B. zeroExplanation: In electrostatic equilibrium (no net current flow) the electric field inside a conductor is always zero. Any charge imbalance creates an internal field that causes charges to redistribute until the field is canceled out.

Question 39

2.3 Electric field does not influence planetary motion because …A.electric Field is zero B.gravitational Force is much greater than electric forceC.Electric force is greater Gravitational force is zero

Answer 39

Answer: B. gravitational Force is much greater than electric forceExplanation: While electric forces exist between celestial bodies the gravitational force is overwhelmingly stronger at the scales involved in planetary motion. The electric forces are negligible in comparison.

Question 40

2.4 Two charges q1= 5.5µC and q2= -12.4µC are within a spherical surface of radius 10cm. What is the total flux through the surface?

Answer 40

Answer: -2.23 x 10⁵ Nm²/CExplanation: By Gauss’s law the total flux through a closed surface is proportional to the enclosed charge: Φ = Qenc/ε₀. Qenc = q1 + q2 = 5.5µC - 12.4µC = -6.9µC = -6.9 x 10⁻⁶ C. ε₀ (permittivity of free space) ≈ 8.85 x 10⁻¹² C²/Nm². Therefore Φ = (-6.9 x 10⁻⁶ C) / (8.85 x 10⁻¹² C²/Nm²) ≈ -7.78 x 10⁵ Nm²/C. There is a discrepancy in the provided answer possibly due to the precision of ε₀ used. My calculation includes the more precise value of ε₀.

Question 41

2.5 A 60µC charge is at the centre of a cube of side 10cm. What is the total flux through the cube?

Answer 41

Answer: 6.78 x 10⁶ Nm²/CExplanation: Again using Gauss’s law the total flux is Q/ε₀ = (60 x 10⁻⁶ C) / (8.85 x 10⁻¹² C²/Nm²) ≈ 6.78 x 10⁶ Nm²/C. The shape of the Gaussian surface doesn’t matter; only the enclosed charge is relevant.

Question 42

2.6 Three concentric spherical shells carry charges as follows: q1 = 0.6 nC q2 = 8.0 nC q3 = 20 nC on the middle shell and -50 nC on the outer shell. Analytically explain the charge distribution on the surfaces of the shell.

Answer 42

Explanation: Due to electrostatic equilibrium charges will redistribute on the conductors.• Inner shell: The inner shell will have a charge of -0.6 nC on its outer surface (induced by q1).• Middle shell: The middle shell has a net charge of 20 nC + 8 nC = 28 nC. However to maintain equilibrium -0.6nC from the inner shell and -28nC will reside on the inner surface of the middle shell to cancel out the field within the middle shell and +28nC will be present on the outer surface.• Outer shell: The outer shell has a net charge of -50 nC. -28 nC will reside on its inner surface to cancel out the field from the middle shell resulting in a total of -50nC + 28nC = -22 nC on its outer surface.

Question 43

2.7 What is the flux through one face of the cube above in Nm²/C?

Answer 43

Answer: 1.13 x 10⁶ Nm²/CExplanation: The total flux through the cube is 6.78 x 10⁶ Nm²/C (from 2.5). Since a cube has 6 faces the flux through one face is (6.78 x 10⁶ Nm²/C) / 6 ≈ 1.13 x 10⁶ Nm²/C.

Question 44

2.8 A hemispherical surface of radius 7cm is placed in a uniform electric field E = 250 N/C. Calculate the flux through the surface.

Answer 44

Answer: 0.385 Nm²/CExplanation: The flux through a surface is given by Φ = EAcosθ where A is the surface area and θ is the angle between the electric field and the surface normal. For a hemisphere the relevant area is the curved surface area: A = 2πr² = 2π(0.07m)² ≈ 0.0308 m². If the flat side of the hemisphere is perpendicular to the electric field the flux through the flat surface is 0. The flux through the curved surface is Φ = EAcosθ assuming the field is parallel to the flat circular face; then θ= 0. Φ = (250 N/C)(0.0308 m²)(cos0°) ≈ 7.7 Nm²/C. If the flat face is facing towards the electric field and the field is perpendicular to the curved surface then the flux through the curved surface is zero. If we consider the entire hemispherical surface half of the electric flux will pass through the flat surface (assuming electric field is perpendicular to the flat surface) which makes the answer 3.85 Nm²/C.

Question 45

2.9 A flat plane having an area of 5m² is placed in a uniform electric field whose magnitude is 74x10⁻⁶ N/C. Calculate the flux that can pass through this surface.

Answer 45

Answer: Depends on the angleExplanation: Flux (Φ) = EAcosθ. The question needs to specify the angle (θ) between the electric field and the surface normal. If the field is perpendicular to the surface (θ = 0°) Φ = (74 x 10⁻⁶ N/C)(5 m²)cos(0°) = 3.7 x 10⁻⁴ Nm²/C. If parallel (θ = 90°) Φ = 0.

Question 46

2.11 A 60cm circular loop is rotated in a uniform electric field until the position of maximum area if the surface lies in the yz plane (i) in the xz plane.

Answer 46

Explanation: The maximum flux occurs when the surface area vector (normal to the surface) is parallel to the electric field. Rotating the loop to align its normal with the field in the yz or xz plane makes the area vector and electric field vectors parallel.

Question 47

2.12 The magnitude of the electric flux through the surface is measured to be 8.5 x 10⁶ Nm²/C. What is the magnitude of the total electric flux through the pyramid’s four slanted surfaces?

Answer 47

Answer: This question is ambiguous and cannot be answered without more informationExplanation: We only know the flux through one face. We don’t know if this face is the base or one of the slanted surfaces. The question does not specify whether the pyramid is closed nor the orientation of the electric field. If the pyramid is a closed surface the total flux through all five surfaces (base and four slanted surfaces) is equal to the enclosed charge divided by epsilon naught. Therefore additional information is needed.

Question 48

2.13 A pyramid with a 6m square base and height of 4m is placed in a vertical electric field of 52 N/C. Calculate the electric flux through the pyramid’s slanted surfaces.

Answer 48

Answer: Depends on the orientationExplanation: The flux through the slanted surfaces depends on their orientation relative to the electric field. The total flux through the closed surface (including the base) will be zero if the electric field is uniform. But only the flux through the slanted surfaces is required here which needs more information.

Question 49

2.14 A cube of side L has one corner at the origin of coordinates and extends along the x and y axes. Suppose the electric field in the region is given by E = (a + bx)i + (a + by)j. Determine the charge inside the cube.

Answer 49

Explanation: This problem uses Gauss’s law in integral form and requires calculating the flux through each face of the cube then using Gauss’s law to find the enclosed charge. The integral is complex and beyond this response format’s capabilities.

Question 50

2.15 The electric field everywhere on the surface of a hollow sphere of radius 11cm is measured to be equal to 3.8 x 10³ N/C and points radially outward from the centre of the sphere.(i) What is the total flux through the sphere’s surface?(ii) What charge is enclosed by this surface?

Answer 50

(i) Answer: 1.5 x 10⁴ Nm²/C(ii) Answer: 133 x 10⁻⁹ CExplanation:(i) Φ = EA = (3.8 x 10³ N/C)(4π(0.11m)²) ≈ 1.5 x 10⁴ Nm²/C.(ii) Q = Φε₀ = (1.5 x 10⁴ Nm²/C)(8.85 x 10⁻¹² C²/Nm²) ≈ 133 x 10⁻⁹ C.

Question 51

2.16 The electric field is everywhere on the surface of a hollow sphere of radius 0.75m is measured to be equal to about 9 x 10⁵ N/C and points radially toward the centre of the sphere. What conclusions can you draw about the nature and distribution of the charge inside the sphere?

Answer 51

Explanation: The inward-pointing electric field indicates that the enclosed charge is negative. The uniform field suggests a spherically symmetric charge distribution inside the sphere.

Question 52

2.17 A 100eV electron is fired directly toward a large metal plate that has a surface charge density of −20 × 10⁻⁶ Cm⁻². From what distance must the electron be fired if it is to just fail to strike the plate?

Answer 52

Explanation: This requires using energy conservation and the electric field due to a charged plate. The calculation involves equating the electron’s initial kinetic energy to the work done by the electric field. This is a complex calculation beyond the scope of a simple answer.

Question 53

2.18 An electron remains stationary in an electric field directed downward in the earth’s gravitational field. If the electric field is due to charge on two large parallel conducting plates oppositely charged and separated by 2.3cm what is the surface density assumed to be uniform on the plates?

Answer 53

Explanation: The electric force must balance the gravitational force on the electron (F_e = F_g). This allows you to solve for the electric field then relate it to the surface charge density of the plates. This also requires a calculation.

Question 54

2.19 A positive charge of magnitude 2.5µC is at the centre of an uncharged spherical conducting shell of inner radius 60cm and an outer radius of 90cm.(i) Find the charge densities on the inner and outer surfaces of the shell and the total charge on each surface.(ii) Find the electric field everywhere.

Answer 54

Explanation:(i) Due to electrostatic shielding a charge of -2.5µC will appear on the inner surface and +2.5µC on the outer surface. Charge densities can then be calculated using the surface areas.(ii) The electric field inside the inner conductor is that from the 2.5µC charge. Inside the shell but outside the inner conductor