Question 1

Q1: 1.1 What is the force between two small charged spheres having charges of $2 	imes 10^{-7} \mathrm{C}$ and $3 	imes 10^{-7} \mathrm{C}$ placed 30 cm...

Accepted Answer

The force between two small charged spheres can be calculated using Coulomb's Law. Given:    Charge on the first sphere, $q_{1} = 2 	imes 10^{-7} \mathrm{C}$    Charge on the second sphere, $q_{2} = 3 	imes 10^{-7} \mathrm{C}$    Distance between the spheres, $r = 30 \mathrm{~cm} = 0.3 \mathrm{~m}...

Question 2

Q10: 1.10 An electric dipole with dipole moment $4 	imes 10^{-9} \mathrm{C} \mathrm{m}$ is aligned at $30^{\circ}$ with the direction of a uniform electri...

Accepted Answer

The torque ($	au$) experienced by an electric dipole in a uniform electric field is given by the cross product of the dipole moment vector ($\mathbf{p}$) and the electric field vector ($\mathbf{E}$). Formula: $$ \mathbf{	au} = \mathbf{p} 	imes \mathbf{E} $$ The magnitude of the torque is given by...

Question 3

Q11: 1.11 A polythene piece rubbed with wool is found to have a negative charge of $3 	imes 10^{-7} \mathrm{C}$. (a) Estimate the number of electrons tran...

Accepted Answer

(a) Number of electrons transferred and direction of transfer This problem uses the principle of quantisation of charge. Given:    Charge on the polythene piece, $q = -3 	imes 10^{-7} \mathrm{C}$    Charge of a single electron, $e = -1.6 	imes 10^{-19} \mathrm{C}$ Formula: The total charge $q$ is...

Question 4

Q12: 1.12 (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...

Accepted Answer

(a) Mutual force of electrostatic repulsion We use Coulomb's Law to find the force between the two spheres. Given:    Charge on sphere A, $q_A = 6.5 	imes 10^{-7} \mathrm{C}$    Charge on sphere B, $q_B = 6.5 	imes 10^{-7} \mathrm{C}$    Distance between centers, $r = 50 \mathrm{~cm} = 0.5 \mathrm...

Question 5

Q13: 1.13 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...

Accepted Answer

The figure shows a uniform electrostatic field directed from a positive plate (top) to a negative plate (bottom). The electric field lines point downwards. 1. Signs of the three charges: The force on a charged particle in an electric field is given by $\mathbf{F} = q\mathbf{E}$.    If the charge $q$...

Question 6

Q14: 1.14 Consider a uniform electric field $\mathbf{E}=3 	imes 10^{3} \hat{\mathbf{i}} \mathrm{N} / \mathrm{C}$. (a) What is the flux of this field throu...

Accepted Answer

The electric flux ($\phi$) through a planar surface is defined as the dot product of the electric field vector ($\mathbf{E}$) and the area vector ($\mathbf{S}$). Formula: $$ \phi = \mathbf{E} \cdot \mathbf{S} = ES \cos	heta $$ where $S$ is the magnitude of the area, and $	heta$ is the angle betwee...

Question 7

Q15: 1.15 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 c...

Accepted Answer

The problem asks for the net electric flux through a closed surface (a cube) placed in a uniform electric field. Given:    Uniform electric field, $\mathbf{E} = 3 	imes 10^3 \hat{\mathbf{i}} \mathrm{~N/C}$.    The surface is a cube of side 20 cm = 0.2 m.    The faces of the cube are parallel to the...

Question 8

Q16: 1.16 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....

Accepted Answer

(a) What is the net charge inside the box? This question can be answered using Gauss's Law. Gauss's Law: The net electric flux ($\phi$) through a closed surface is directly proportional to the net charge ($q_{net}$) enclosed by that surface. $$ \phi = \frac{q_{net}}{\varepsilon_0} $$ We can rearrang...

Question 9

Q17: 1.17 A point charge $+10 \mu \mathrm{C}$ 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 m...

Accepted Answer

This problem can be solved by applying Gauss's Law and using symmetry arguments. Given:    Point charge, $q = +10 \mu \mathrm{C} = 10 	imes 10^{-6} \mathrm{C}$    Side of the square, $s = 10 \mathrm{~cm} = 0.1 \mathrm{~m}$ *   Distance of the charge from the center of the square, $d = 5 \mathrm{~cm...

Question 10

Q18: 1.18 A point charge of $2.0 \mu \mathrm{C}$ is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the sur...

Accepted Answer

This problem is a direct application of Gauss's Law. Gauss's Law: The net electric flux ($\phi_{net}$) through any closed surface is equal to the net charge enclosed ($q_{enclosed}$) by the surface, divided by the permittivity of free space ($\varepsilon_0$). $$ \phi_{net} = \frac{q_{enclosed}}{\var...

Question 11

Q19: 1.19 A point charge causes an electric flux of $-1.0 	imes 10^{3} \mathrm{Nm}^{2} / \mathrm{C}$ to pass through a spherical Gaussian surface of 10.0...

Accepted Answer

(a) Flux if the radius of the Gaussian surface were doubled According to Gauss's Law, the total electric flux through a closed surface depends only on the net charge enclosed by the surface. It is independent of the size or shape of the Gaussian surface. $$ \phi = \frac{q_{enclosed}}{\varepsilon_0}...

Question 12

Q2: 1.2 The electrostatic force on a small sphere of charge $0.4 \mu \mathrm{C}$ due to another small sphere of charge $-0.8 \mu \mathrm{C}$ in air is 0.2...

Accepted Answer

This problem involves applying Coulomb's Law and Newton's third law of motion. Given:    Charge on the first sphere, $q_{1} = 0.4 \mu \mathrm{C} = 0.4 	imes 10^{-6} \mathrm{C}$    Charge on the second sphere, $q_{2} = -0.8 \mu \mathrm{C} = -0.8 	imes 10^{-6} \mathrm{C}$    Electrostatic force on t...

Question 13

Q20: 1.20 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 	imes 10^{3} \mathr...

Accepted Answer

For a point outside a charged conducting sphere, the electric field is the same as if the entire charge of the sphere were concentrated at its center. Formula: The magnitude of the electric field ($E$) at a distance ($r$) from the center of a sphere with charge ($q$) is given by: $$ E = k \frac{|q|}...

Question 14

Q21: 1.21 A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of $80.0 \mu \mathrm{C} / \mathrm{m}^{2}$. (a) Find the char...

Accepted Answer

(a) Find the charge on the sphere. Surface charge density ($\sigma$) is defined as the charge per unit surface area. Formula: $$ \sigma = \frac{q}{A} $$ where $q$ is the total charge and $A$ is the surface area of the sphere. The surface area of a sphere is $A = 4\pi r^2$. We can find the charge $q$...

Question 15

Q22: 1.22 An infinite line charge produces a field of $9 	imes 10^{4} \mathrm{N} / \mathrm{C}$ at a distance of 2 cm . Calculate the linear charge density...

Accepted Answer

The electric field produced by an infinite line charge is given by a specific formula derived from Gauss's Law. Formula: The magnitude of the electric field ($E$) at a perpendicular distance ($r$) from an infinite line charge with uniform linear charge density ($\lambda$) is: $$ E = \frac{\lambda}{2...

Question 16

Q23: 1.23 Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite sig...

Accepted Answer

Let's denote the first plate as Plate 1 and the second plate as Plate 2. Let Plate 1 have a positive surface charge density, $\sigma = +17.0 	imes 10^{-22} \mathrm{C} / \mathrm{m}^{2}$, and Plate 2 have a negative surface charge density, $-\sigma = -17.0 	imes 10^{-22} \mathrm{C} / \mathrm{m}^{2}$...

Question 17

Q3: 1.3 Check that the ratio $k e^{2} / G m_{e} m_{p}$ is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What...

Accepted Answer

1. Checking for Dimensionlessness To check if the ratio is dimensionless, we will analyze the units of each constant in the expression.    $k$ (Coulomb's constant): $\mathrm{N} \mathrm{m}^{2} \mathrm{C}^{-2}$    $e$ (elementary charge): $\mathrm{C}$    $G$ (Gravitational constant): $\mathrm{N} \math...

Question 18

Q4: 1.4 (a) Explain the meaning of the statement 'electric charge of a body is quantised'. (b) Why can one ignore quantisation of electric charge when dea...

Accepted Answer

(a) Meaning of 'electric charge of a body is quantised' The statement 'electric charge of a body is quantised' means that electric charge is not a continuous quantity but occurs in discrete packets or units. All free charges are integral multiples of a basic, fundamental unit of charge, denoted by '...

Question 19

Q5: 1.5 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 ho...

Accepted Answer

The law of conservation of charge states that the total electric charge in an isolated system remains constant. Charge can neither be created nor destroyed; it can only be transferred from one body to another. The observation of charges appearing on a glass rod and a silk cloth when they are rubbed...

Question 20

Q6: 1.6 Four point charges $q_{\mathrm{A}}=2 \mu \mathrm{C}, q_{\mathrm{B}}=-5 \mu \mathrm{C}, q_{\mathrm{C}}=2 \mu \mathrm{C}$, and $q_{\mathrm{D}}=-5 \m...

Accepted Answer

Let us analyze the forces exerted on the $1 \mu \mathrm{C}$ charge (let us call it $q_O$) placed at the center O of the square ABCD. Given:    Charge at corner A, $q_A = 2 \mu \mathrm{C}$    Charge at corner B, $q_B = -5 \mu \mathrm{C}$    Charge at corner C, $q_C = 2 \mu \mathrm{C}$    Charge at co...

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