Electric Charge
A fundamental property of matter, characterized by two distinct types: positive and negative.
Like charges repel, unlike charges attract.
Measured in coulombs (C).
Quantized: exists in discrete units of 1.6 × 10⁻¹⁹ C.
Coulomb’s Law
The force between two point charges is proportional to the product of their charges divided by the square of the distance between them.
Mathematically: F = (1/4πε₀) * (q₁q₂/r²)
Electric Field
Field intensity at a point
A vector quantity, visualized by field lines.
Lines originate from positive charges and terminate on negative charges.
The electric field at a distance r from a point charge q is:
E = (1/4πε₀) * (q/r²)
Electric Dipole
A pair of charges, equal in magnitude but opposite in sign, separated by a short distance.
Electric field at a distance is proportional to the dipole moment (p = qd).
Experiences a torque in an electric field: τ = p × E
Electric Flux
Measures the number of electric field lines passing through a surface.
Gauss’s Law: Total electric flux through a closed surface equals the net enclosed charge divided by ε₀.
This chapter lays the groundwork for understanding electrostatics, essential for studying electromagnetism and related fields.
1 What is the force between two small charged spheres having charges of 2 × 10-7 and 3 × 10-7 C placed 30 cm apart in air?
Ans :Given:
Charge on the 1st sphere,
q₁ = 2 x 10⁻⁷ C
Charge on the 2nd sphere,
q₂ = 3 x 10⁻⁷ C
Distance between the spheres,
r = 30 cm = 0.3 m
The electrostatic force (F) between the two charged spheres can be calculated using Coulomb’s Law:
F = (1 / 4πε₀) * (q₁q₂ / r²)
Where:
ε₀ is the permittivity of free space
1 / 4πε₀
= 9 x 10⁹ N m²/C²
Substituting the values:
F = (9 x 10⁹) * ((2 x 10⁻⁷) * (3 x 10⁻⁷)) / (0.3)²
F = 6 x 10⁻³ N
2 The electrostatic force on a small sphere of charge 0.4 mC due to another small sphere of charge –0.8 mC in air is 0.2 N.
(a) What is the distance between the two spheres?
(b) What is the force on the second sphere due to the first?
Ans :(a) Given,
F denote Electrostatic force on the first sphere,
= 0.2 N
Charge on the first sphere, q₁ = 0.4 μC = 0.4 × 10⁻⁶ C
Charge on the second sphere, q₂ = -0.8 μC = -0.8 × 10⁻⁶ C
The electrostatic force between the spheres is given by Coulomb’s law:
F = (1 / 4πε₀) * (q₁q₂ / r²)
Where:
ε₀ is the permittivity of free space
1 / 4πε₀ = 9 × 10⁹ N m²/C²
Rearranging the equation to solve for r:
r² = (1 / 4πε₀) * (q₁q₂ / F)
Substituting the values:
r² = (9 × 10⁹) * ((0.4 × 10⁻⁶) * (-0.8 × 10⁻⁶)) / 0.2
r² = 144 × 10⁻⁴
r = √(144 × 10⁻⁴) = 0.12 m
Therefore, the distance between the two spheres is 0.12 meters.
(b) According to Newton’s Third Law of Motion, the force exerted by the first sphere on the second sphere is equal in magnitude and opposite in direction to the force exerted by the second sphere on the first. Hence, the force on the second sphere due to the first is also 0.2 N.
3 Check that the ratio ke2/Gmemp is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?
Ans :Checking Dimensionality
Let’s analyze the dimensions of each term in the ratio:
ke²:
Coulomb’s constant, k, has units of newton-meter squared per coulomb squared(N m²/C².)
he elementary charge, e, has units of Coulombs C.
So, ke² has dimensions of N m².
Gmemp:
G (gravitational constant) has dimensions of N m²/kg².
me and mp (masses of electron and proton) have dimensions of kg.
So, Gmemp has dimensions of N m².
As we can see, both the numerator and denominator have the same dimensions of N m². Therefore, their ratio, ke²/Gmemp, is dimensionless.
Calculating the Ratio
To calculate the value of this ratio, we need to look up the values of the constants involved:
k denote Coulomb’s constant
= 8.98755 × 10⁹ N m²/C²
e denote elementary charge
= 1.60218 × 10⁻¹⁹ C
G denote gravitational constant
=6.67430 × 10⁻¹¹ N m²/kg²
me (electron mass): 9.10938 × 10⁻³¹ kg
mp (proton mass): 1.67262 × 10⁻²⁷ kg
susbituting values into the ratio,
ke²/Gmemp = (8.98755 × 10⁹ N m²/C²) × (1.60218 × 10⁻¹⁹ C)² / [(6.67430 × 10⁻¹¹ N m²/kg²) × (9.10938 × 10⁻³¹ kg) × (1.67262 × 10⁻²⁷ kg)]
After calculating, we find that:
ke²/Gmemp ≈ 2.27 × 10³⁹
Significance of the Ratio
This dimensionless ratio, approximately equal to 2.27 × 10³⁹, represents the relative strength of the electromagnetic force to the gravitational force between an electron and a proton. The substantial magnitude of the ratio highlights the dominance of the electromagnetic force over the gravitational force in atomic interactions. This is why gravitational forces are negligible in most atomic and molecular phenomena, and electromagnetic forces dominate.
4 (a) Explain the meaning of the statement ‘electric charge of a body is quantised’.
(b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges?
Ans :(a) The principle of charge quantization asserts that electric charge is not infinitely divisible but rather exists in discrete units. Specifically, the charge on any object is an integer multiple of the elementary charge, denoted by e.
(b) When dealing with macroscopic objects, the quantization of charge becomes negligible due to the vast number of charges involved. Consequently, it is permissible to treat charge as a continuous variable in such scenarios.
5. When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.
Ans :Conservation of Charge in Electrostatic Charging
The law of conservation of charge states that the net electric charge of an isolated system remains constant. When a glass rod is rubbed with a silk cloth, the friction between them causes electrons to transfer from one material to the other. This transfer of electrons results in a charge separation.
Glass Rod: Loses electrons, becoming positively charged.
Silk Cloth: Gains electrons, becoming negatively charged.
While charges are created on both objects, the total charge of the system (glass rod + silk cloth) remains constant. The negative charge gained by the silk cloth is equal in magnitude to the positive charge gained by the glass rod. Thus, the net charge of the isolated system is zero both before and after the rubbing process.
In essence, the law of conservation of charge is upheld because:
Charge is quantized: It exists in discrete units (electrons and protons).
Charge is conserved: The total charge of an isolated system remains constant.
Therefore, the phenomenon of electrostatic charging, where charges appear on both objects, is entirely consistent with the law of conservation of charge.
6. Four point charges qA = 2 mC, qB = –5 mC, qC = 2 mC, and qD = –5 mC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 mC placed at the centre of the square?
Ans :Analyzing the Forces on the Center Charge
Understanding the Problem:
We have a square with four charges at its corners:
qA = 2 mC
qB = -5 mC
qC = 2 mC
qD = -5 mC
A charge of 1 mC is placed at the center of the square. We need to find the net force on this center charge.
Key Points:
Symmetry: Due to the symmetry of the square and the arrangement of charges, the forces from charges A and C will cancel each other out. Similarly, the forces from charges B and D will cancel each other out.
Coulomb’s Law: This law governs the electrostatic force between charged particles.
Given,
The electrostatic force of attraction or repulsion between two point charges, q1 and q2, separated by a distance r
F = k * (q1 * q2)
———————–
r^2
where k is Coulomb’s constant.
Solution:
Due to the symmetry of the charge distribution, the forces exerted on the center charge by opposite corner charges cancel each other, resulting in a net force of 0.
Therefore, the force on the 1 mC charge at the center of the square is 0 N.
7. (a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?
(b) Explain why two field lines never cross each other at any point?
Ans :(a) An electrostatic field line is a continuous curve because it represents the path that a positive test charge would follow if placed at that point. If a field line were to suddenly break, it would imply that the test charge would spontaneously change its direction of motion without any external force acting on it, which is not physically possible. Therefore, field lines must be continuous to represent the smooth flow of a test charge in an electric field.
(b) Two field lines cannot cross each other at any point because at any point in an electric field, there can only be one unique direction of the electric field. If two field lines were to cross, it would imply that at that point, the electric field has two different directions, which is contradictory. Hence, field lines must never intersect to maintain the uniqueness of the electric field direction at every point.
8. Two point charges qA = 3 mC and qB = –3 mC are located 20 cm apart in vacuum.
(a) What is the electric field at the midpoint O of the line AB joining the two charges?
(b) If a negative test charge of magnitude 1.5 × 10–9 C is placed at this point, what is the force experienced by the test charge?
Ans :(a) To find the electric field at the midpoint O, we need to calculate the electric field due to each charge and then add them vectorially.
Electric field due to qA:
Distance from qA to O: r = 10 cm = 0.1 m
Electric field magnitude: E_A
= k * |qA| / r^2
Direction: Towards qB (since qA is positive)
Electric field due to qB:
Distance from qB to O: r
= 0.1 m
Electric field magnitude: E_B = k * |qB| / r^2
Direction: Towards qB (since qB is negative)
Since both electric fields point in the same direction (towards qB), the net electric field at O is the sum of their magnitudes:
E_net = E_A + E_B = k * |qA| / r^2 + k * |qB| / r^2
Substituting the values:
E_net = (9 * 10^9 N m^2/C^2) * (3 * 10^-3 C / (0.1 m)^2 + 3 * 10^-3 C / (0.1 m)^2)
E_net = 5.4 * 10^9 N/C
Therefore, the electric field at the midpoint O is 5.4 x 10^9 N/C, directed towards qB.
(b) To find the force on the test charge, we use the formula:
F = q * E
where q is the test charge and E is the electric field at that point.
Substituting the values:
F = (1.5 * 10^-9 C) * (5.4 * 10^9 N/C)
F = -8.1 N
Hence, the force experienced by the test charge is 8.1 N, directed towards qA (opposite to the direction of the electric field).
9. A system has two charges qA = 2.5 × 10–7 C and qB = –2.5 × 10–7 C located at points A: (0, 0, –15 cm) and B: (0,0, +15 cm), respectively. What are the total charge and electric dipole moment of the system?
Ans :Total Charge and Electric Dipole Moment
Total Charge
The total charge of the system is the algebraic sum of the individual charges:
Q_total = qA + qB = 2.5 * 10^(-7) C + (-2.5 * 10^(-7) C) = 0 C
Therefore, the total charge of the system is 0 C.
Electric Dipole Moment
p = q * d
where:
q is the magnitude of each charge
d is the distance between the charges
In this case:
q = 2.5 * 10^(-7) C
d = 30 cm = 0.3 m
So, the electric dipole moment is:
p = (2.5 * 10^(-7) C) * (0.3 m) = 7.5 * 10^(-8) C*m
Therefore, the electric dipole moment of the system is 7.5 * 10^(-8) C*m.
The direction of the dipole moment is from the negative charge to the positive charge, which is along the positive z-axis in this case.
10. An electric dipole with dipole moment 4 × 10–9 C m is aligned at 30° with the direction of a uniform electric field of magnitude 5 × 104 NC–1 . Calculate the magnitude of the torque acting on the dipole.
Ans :The torque (τ) on an electric dipole in a uniform electric field (E) is given by:
τ = pE sinθ
where:
p = dipole moment
E = electric field strength
θ = angle formed by the dipole moment vector and the electric field vector, measured in radians or degrees
Given:
p = 4 × 10^(-9) C m
E = 5 × 10^4 N/C
θ = 30°
Substituting the values:
τ =
(4 × 10^(-9) C m) × (5 × 10^4 N/C) × sin(30°)
τ = (20 × 10^(-5)) × (1/2) N m
τ = 10^(-4) N m
Therefore, the magnitude of the torque acting on the dipole is 10^(-4) N m.
11. A polythene piece rubbed with wool is found to have a negative charge of 3 × 10–7 C.
(a) Estimate the number of electrons transferred (from which to which?)
(b) Is there a transfer of mass from wool to polythene?
Ans :(a) To find the number of electrons transferred, we can use the following formula:
Number of electrons =
Total charge
———————-
Charge of one electron
Given:
Total charge = 3 × 10^-7 C
Charge of one electron = 1.6 × 10^-19 C
Number of electrons =
(3 × 10^-7 C)
——————
(1.6 × 10^-19 C)
≈ 1.875 × 10^12 electrons
Since the polythene piece has a negative charge, it has gained electrons. Therefore, 1.875 × 10^12 electrons were transferred from the wool to the polythene.
(b) Yes, a small amount of mass is transferred from the wool to the polythene during the charging process due to the transfer of electrons.. Each electron has a mass of approximately 9.11 × 10^-31 kg. So, the total mass transferred is:
Mass transferred = Number of electrons × Mass of one electron
Mass transferred ≈ (1.875 × 10^12) × (9.11 × 10^-31 kg)
≈ 1.7 × 10^-18 kg
However, this mass transfer is extremely small and practically negligible.
12. (a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10–7 C? The radii of A and B are negligible compared to the distance of separation.
(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?
Ans :(a)The electrostatic force between the two charged spheres can be calculated using Coulomb’s law, which states:
F =
k * (q1 * q2)
———————
r^2
where k is Coulomb’s constant (9 × 10^9 N m²/C²), q1 and q2 are the charges on the spheres (both 6.5 × 10^-7 C), and r is the distance between their centers (0.5 m). Substituting these values into the formula and calculating, we find that the force of repulsion between the spheres is approximately 1.52 × 10^-2 N.
(b) If each sphere is charged double the above amount, the new charges are:
q1′ = q2′ = 2 * (6.5 x 10^-7 C) = 1.3 x 10^-6 C
The distance between the spheres is halved:
r’ = 0.5 m / 2 = 0.25 m
Using Coulomb’s law again:
F’ = (9 x 10^9 N m^2/C^2) * (1.3 x 10^-6 C)^2 / (0.25 m)^2
Calculating this gives:
F’ ≈ 0.24 N
Therefore, the force of repulsion when the charges are doubled and the distance is halved is approximately 0.24 N.
Figure 1.30 shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?
Ans :Given:
A diagram showing the tracks of three charged particles in a uniform electrostatic field.
Analysis:
Signs of Charges:
Particle 1 and 2: Both move towards the positively charged plate and away from the negatively charged plate. This indicates that they are negatively charged.
Particle 3: The negatively charged plate attracts the particle, while the positively charged plate repels it.. This indicates that it is positively charged.
Charge-to-Mass Ratio (q/m):
The displacement or deflection of a charged particle in a uniform electric field is directly proportional to its charge-to-mass ratio, assuming a constant velocity.
Particle 3: It shows the maximum deflection among the three particles.
Hence, it has the highest charge-to-mass ratio.
Conclusion:
Particles 1 and 2 are negatively charged.
Particle 3 is positively charged.
Particle 3 has the highest charge-to-mass ratio.
14. Consider a uniform electric field E = 3 × 103 î N/C.
(a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane?
(b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis?
Ans : (a) Flux through a surface is given by:
Φ = ∫ E.dA
Since the electric field is uniform and perpendicular to the plane of the square, the flux simplifies to:
Φ = EA
Where:
* E = 3 × 10³ N/C
* A = (10 cm)² = 0.01 m²
So, Φ = (3 × 10³) N/C × 0.01 m² = 30 Nm²/C
(b) If the normal to the plane is inclined at 60° to the x-axis, the perpendicular component of the electric field is:
E_perpendicular = E cos(60°) = (3 × 10³) N/C × 0.5 = 1.5 × 10³ N/C
Now, the flux is:
Φ = E_perpendicular × A = (1.5 × 10³) N/C × 0.01 m² = 15 Nm²/C
15. What is the net flux of the uniform electric field of Exercise 1.14 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?
Ans : In a uniform electric field, the net flux through a closed surface is always zero. This happens because of Gauss’s Law..
Therefore, the net flux through the cube in this case is zero.
16. Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 × 103 Nm2 /C.
(a) What is the net charge inside the box?
(b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or Why not?
Ans :Given:
Net outward flux through the surface of the box (Φ) = 8.0 x 10³ Nm²/C
(ε₀) denote Permittivity of free space
= 8.854 x 10⁻¹² C²/Nm²
Solution:
Part (a):
Apply Gauss’s Law:
Gauss’s Law states that the net electric flux through a closed surface is equal to the net charge enclosed (q) divided by the permittivity of free space (ε₀):
Φ = q / ε₀
Calculate the Net Charge:
Rearranging the formula to find the net charge:
q = Φ * ε₀
Substitute Values:
q = (8.0 x 10³ Nm²/C) * (8.854 x 10⁻¹² C²/Nm²)
Calculate the Net Charge:
q ≈ 7.08 x 10⁻⁸ C = 0.07 μC
Answer:
Therefore, the net charge inside the box is approximately 0.07 μC.
Part (b):
No.
The net electric flux piercing out through a body depends solely on the net charge contained within the body. If the net flux is zero, it implies that the net charge inside the body is also zero. This does not necessarily mean that the body contains no charge. It could have equal amounts of positive and negative charges, resulting in a net charge of zero.
17. A point charge +10 mC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.31. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.)
Ans :Given:
(q) denote Point charge
= 10 μC
= 10 x 10^-6 C
(a) denote Edge length of the cube
= 10 cm
= 0.1 m
Permittivity of free space (ε₀) = 8.854 x 10^-12 C^2/Nm^2
Solution:
Apply Gauss’s Law:
Gauss’s Law states that the total electric flux through a closed surface is directly proportional to the net charge enclosed by it, and the constant of proportionality is the permittivity of free space.
Φ_total = q / ε₀
Calculate Flux through One Face:
Since the cube has six faces, the flux through each face (Φ) is one-sixth of the total flux:
Φ = Φ_total / 6 = q / (6ε₀)
Substitute Values:
Φ = (10 x 10^-6 C) / (6 * 8.854 x 10^-12 C^2/Nm^2)
Calculate Flux:
Φ ≈ 1.88 x 10^5 Nm^2/C
Answer:
hence, the electric flux through the square is about 1.88 x 10^5 Nm^2/C.
18. A point charge of 2.0 mC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?
Ans :Given:
Charge enclosed (Q)
= 2.0 mC
= 2.0 × 10⁻³ C
(ε₀) denote Permittivity of free space
≈ 8.85 × 10⁻¹² C²/Nm²
Using Gauss’s Law:
Electric Flux (Φ) = Q / ε₀
Substituting the values:
Φ = (2.0 × 10⁻³ C) / (8.85 × 10⁻¹² C²/Nm²)
Calculating this gives:
Φ ≈ 2.26 × 10⁸ Nm²/C
Therefore, the net electric flux through the surface is approximately 2.26 × 10⁸ Nm²/C.
19. A point charge causes an electric flux of –1.0 × 103 Nm2 /C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge.
(a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?
(b) What is the value of the point charge?
Ans :Given:
Electric flux (Φ)
= -1.0 × 10³ Nm²/C
Radius of the Gaussian surface (r)
= 10.0 cm
= 0.1 m
To find:
(a) Electric flux through the surface if the radius is doubled
(b) Value of the point charge
Solution
(a) Electric Flux through the Doubled Radius Surface
According to Gauss’s Law, the electric flux through a closed surface depends only on the net charge enclosed by the surface, not on the size or shape of the surface.
Therefore, if the radius of the Gaussian surface is doubled, the electric flux will remain the same.
So, the electric flux through the doubled radius surface is still -1.0 × 10³ Nm²/C.
(b) Value of the Point Charge
Gauss’s Law states:
Φ = Q / ε₀
where:
Φ = electric flux
Q = net charge enclosed
ε₀ = permittivity of free space
(approximately 8.85 × 10⁻¹² C²/Nm²)
Rearranging the equation to solve for Q:
Q = Φ * ε₀
Substituting the given values:
Q = (-1.0 × 10³ Nm²/C) * (8.85 × 10⁻¹² C²/Nm²)
Calculating this gives:
Q ≈ -8.85 × 10⁻⁹ C
Therefore, the value of the point charge is approximately -8.85 × 10⁻⁹ C.
20. A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 103 N/C and points radially inward, what is the net charge on the sphere?
Ans :Understanding the Problem:
We have a conducting sphere with an unknown charge. We are given the electric field at a distance of 20 cm from the center of the sphere and asked to find the net charge on the sphere.
Solution:
1. Using Gauss’s Law:
For a conducting sphere, the electric field outside the sphere behaves as if all the charge were concentrated at its center. This allows us to use Gauss’s Law, which states that the electric flux through a closed surface is equal to the net charge enclosed divided by the permittivity of free space (ε₀).
2. Applying Gauss’s Law:
We can imagine a spherical Gaussian surface centered at the sphere’s center with a radius of 20 cm (0.2 m). The electric field vector is uniform and orthogonal to the surface at all points.
Electric Flux (Φ) =
E * A = Q
———-
ε₀
where:
E = electric field
A = surface area of the Gaussian sphere (4πr²)
Q = net charge enclosed
ε₀ = permittivity of free space (approximately 8.85 × 10⁻¹² C²/Nm²)
3. Solving for Q:
Q = E * A * ε₀
Substituting the given values:
Q = (1.5 × 10³ N/C) * (4π * (0.2 m)²) * (8.85 × 10⁻¹² C²/Nm²)
Calculating this gives:
Q ≈ 6.67 × 10⁻⁸ C
Hence, the net charge on the sphere is about 6.67 x 10^-8 Coulombs.
21. A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 mC/m2 .
(a) Find the charge on the sphere.
(b) What is the total electric flux leaving the surface of the sphere?
Ans :Given:
D denote Diameter of the sphere
= 2.4 m
Surface charge density, σ = 80.0 mC/m² = 80.0 × 10⁻⁶ C/m²
(a) Finding the Charge on the Sphere
1. Calculate the radius:
Radius, r = D/2 = 1.2 m
2. Calculate the surface area:
Surface area, A = 4πr² = 4π(1.2)² m²
3. Calculate the charge:
Charge, Q = σA = (80.0 × 10⁻⁶ C/m²) × 4π(1.2)² m² ≈ 1.45 × 10⁻³ C
Therefore, the charge on the sphere is approximately 1.45 × 10⁻³ C.
(b)Determine Total Electric Flux Leaving the Surface
According to Gauss’s Law, the total electric flux (Φ) through a closed surface is equal to the net charge enclosed (Q) divided by the permittivity of free space (ε₀):
Φ = Q / ε₀
where:
ε₀ = 8.85 × 10⁻¹² C²/Nm² (permittivity of free space)
Substituting the values:
Φ = (1.45 × 10⁻³ C) / (8.85 × 10⁻¹² C²/Nm²) ≈ 1.64 × 10⁸ Nm²/C
Therefore, the total electric flux leaving the surface of the sphere is approximately 1.64 × 10⁸ Nm²/C.
22. An infinite line charge produces a field of 9 × 104 N/C at a distance of 2 cm. Calculate the linear charge density.
Ans :Given:
(E) denote Electric field
= 9 x 10⁴ N/C
(d) denote Distance from the line charge
= 2 cm
= 0.02 m
Permittivity of free space (ε₀) = 1 / (4π * 9 x 10⁹ N m² C⁻²)
Formula:
Electric field from an infinite line charge with linear charge density λ at distance d is:
E = λ / (2πε₀d)
Calculation:
Rearranging the formula to find λ:
λ = 2πε₀dE
Substituting the given values:
λ = 2π * (1 / (4π * 9 x 10⁹)) * 0.02 * 9 x 10⁴
Simplifying:
λ = 10 x 10⁻⁶ C/m
Answer:
hence, the linear charge density of the infinite line charge is 10 µC/m.
23. Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 × 10–22 C/m2 . What is E:
(a) in the outer region of the first plate,
(b) in the outer region of the second plate, and
(c) between the plates?
Ans :Given:
Surface charge density, σ = 17.0 × 10^(-22) C/m²
(a)
For a large, thin, uniformly charged plate, the electric field at a point outside the plate is given by:
E = σ / (2 * ε₀)
where:
ε₀ is the permittivity of free space (approximately 8.85 × 10^(-12) C²/Nm²)
Substituting the given values:
E = (17.0 × 10^(-22) C/m²) / (2 * 8.85 × 10^(-12) C²/Nm²) ≈ 9.60 × 10^(-11) N/C
Therefore, the electric field in the outer region of the first plate is approximately 9.60 × 10^(-11) N/C.
(b)
Since the second plate is identical to the first plate in terms of charge density and size, the electric field in its outer region will also be the same.
Therefore, the electric field in the outer region of the second plate is also approximately 9.60 × 10^(-11) N/C.
(c)
The electric fields created by both plates combine within the plates. As a result, the total electric field between the plates is:
E = σ / ε₀
Substituting the values:
E = (17.0 × 10^(-22) C/m²) / (8.85 × 10^(-12) C²/Nm²)
≈ 1.92 × 10^(-10) N/C
Therefore, the electric field between the plates is approximately 1.92 × 10^(-10) N/C.
