Rbse Solution for Class 12 Physics Chapter 1: Electric Charges and Fields

Rbse Solution for Class 12 Physics Chapter 1: Electric Charges and Fields

Chapter 1

(Open)

1.1 What is the force between two small charged spheres having charges of $2 \times 10^{-7}\text{ C}$ and $3 \times 10^{-7}\text{ C}$ placed $30\text{ cm}$ apart in air?

This problem requires the application of Coulomb’s Law:

$$F = k \frac{|q_1 q_2|}{r^2}$$

• $q_1 = 2 \times 10^{-7}\text{ C}$
• $q_2 = 3 \times 10^{-7}\text{ C}$
• $r = 30\text{ cm} = 0.30\text{ m}$
• $k = 9 \times 10^9\text{ N m}^2\text{ C}^{-2}$

$$F = (9 \times 10^9) \frac{|(2 \times 10^{-7})(3 \times 10^{-7})|}{(0.30)^2}\text{ N}$$

$$F = (9 \times 10^9) \frac{6 \times 10^{-14}}{0.09}\text{ N}$$

$$F = 6 \times 10^{-3}\text{ N}$$

The force is $6 \times 10^{-3}\text{ N}$ (repulsive, since both charges are positive).

1.2 The electrostatic force on a small sphere of charge $0.4\text{ $\mu$C}$ due to another small sphere of charge $-0.8\text{ $\mu$C}$ in air is $0.2\text{ N}$. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first?

• $q_1 = 0.4\text{ $\mu$C} = 0.4 \times 10^{-6}\text{ C}$
• $q_2 = -0.8\text{ $\mu$C} = -0.8 \times 10^{-6}\text{ C}$
• $F = 0.2\text{ N}$

(a) What is the distance between the two spheres?

Rearranging Coulomb’s Law: $r^2 = k \frac{|q_1 q_2|}{F}$

$$r^2 = (9 \times 10^9) \frac{|(0.4 \times 10^{-6}) (-0.8 \times 10^{-6})|}{0.2}\text{ m}^2$$

$$r^2 = (9 \times 10^9) \frac{3.2 \times 10^{-13}}{0.2}\text{ m}^2 = 144 \times 10^{-4}\text{ m}^2$$

$$r = \sqrt{144 \times 10^{-4}}\text{ m} = 12 \times 10^{-2}\text{ m} = 0.12\text{ m}$$

The distance is $0.12\text{ m}$.

(b) What is the force on the second sphere due to the first?

By Newton’s Third Law (Action-Reaction pair), the force exerted by the first sphere on the second is equal in magnitude and opposite in direction to the force exerted by the second sphere on the first. The magnitude of the force is given.

The force is $0.2\text{ N}$ (attractive, as the charges are opposite).

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 does the ratio signify?

• Checking Dimensions:$$\left[ \frac{k e^2}{G m_e m_p} \right] = \frac{([\text{N m}^2 \text{C}^{-2}])([\text{C}^2])}{([\text{N m}^2 \text{kg}^{-2}])([\text{kg}])([\text{kg}])} = \frac{\text{N m}^2}{\text{N m}^2} = 1$$The ratio is dimensionless.
• Value of the Ratio:Using constants: $k \approx 9 \times 10^9$, $e \approx 1.602 \times 10^{-19}$, $G \approx 6.67 \times 10^{-11}$, $m_e \approx 9.11 \times 10^{-31}$, $m_p \approx 1.67 \times 10^{-27}$.$$\text{Value} \approx \frac{(9 \times 10^9)(1.602 \times 10^{-19})^2}{(6.67 \times 10^{-11})(9.11 \times 10^{-31})(1.67 \times 10^{-27})} \approx 2.27 \times 10^{39}$$
• Significance:The ratio represents the ratio of the electrostatic force to the gravitational force between an electron and a proton. Its immense value ($2.27 \times 10^{39}$) signifies that the electrostatic force is vastly stronger than the gravitational force at the atomic level.

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 dealing with macroscopic i.e., large scale charges?

(a) Meaning of Quantisation of Electric Charge

Quantisation of electric charge means that the charge ($q$) on any object is always an integral multiple of the basic unit of charge, $e$ (the elementary charge, $\approx 1.602 \times 10^{-19}\text{ C}$).

$$q = n e \quad (\text{where } n = 0, \pm 1, \pm 2, \dots)$$

Charge is not continuous but exists in discrete “packets.”

(b) Ignoring Quantisation for Macroscopic Charges

For macroscopic charges (large scale), the amount of charge involved is very large compared to $e$. For example, $1\text{ $\mu$C}$ contains $n \approx 6.25 \times 10^{12}$ elementary charges.

Since the number of elementary charges is extremely large, the change in charge appears continuous. We treat charge as a continuous variable for large-scale charges because the difference between $n$ and $n+1$ units of charge is practically undetectable.

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 how this observation is consistent with the law of conservation of charge.

The Law of Conservation of Charge states that the net charge of an isolated system remains constant.

• Initial State: Both the glass rod and the silk cloth are electrically neutral. Net charge of the system $= 0$.
• Process: When rubbed, electrons are transferred from the glass rod (becoming positive) to the silk cloth (becoming negative). No charge is created or destroyed.
• Final State: The positive charge ($+Q$) on the rod is exactly equal in magnitude to the negative charge ($-Q$) on the silk.
• Net Charge: The net charge of the system remains $(+Q) + (-Q) = 0$.

Since charge is only transferred from one body to the other, the observation is entirely consistent with the law of conservation of charge.

1.6 Four point charges $q_A = 2\text{ $\mu$C}$, $q_B = -5\text{ $\mu$C}$, $q_C = 2\text{ $\mu$C}$, and $q_D = -5\text{ $\mu$C}$ are located at the corners of a square ABCD of side $10\text{ cm}$. What is the force on a charge of $1\text{ $\mu$C}$ placed at the centre of the square?

Let $q_0 = 1\text{ $\mu$C}$ be the charge at the centre $O$.

The distance $r$ from the centre $O$ to any corner is the same.

The net force $\vec{F}$ on $q_0$ is the vector sum of the forces due to the four corner charges: $\vec{F} = \vec{F}_A + \vec{F}_B + \vec{F}_C + \vec{F}_D$.

1. Forces $\vec{F}_A$ and $\vec{F}_C$:Since $q_A = q_C$ and the distance $r$ is the same, $F_A = F_C$. Both $q_A$ and $q_C$ are positive, so their forces on $q_0$ are repulsive and opposite along the diagonal $AC$.$$\vec{F}_A + \vec{F}_C = 0$$
2. Forces $\vec{F}_B$ and $\vec{F}_D$:Since $q_B = q_D$ and the distance $r$ is the same, $F_B = F_D$. Both $q_B$ and $q_D$ are negative, so their forces on $q_0$ are attractive and opposite along the diagonal $BD$.$$\vec{F}_B + \vec{F}_D = 0$$

The net force on the charge $q_0$ is $\vec{F} = 0 + 0 = 0$.

The force on the charge of $1\text{ $\mu$C}$ placed at the centre is zero.

1.7 (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?

(a) Continuity of Field Lines

An electric field line represents the path a small positive test charge would follow. The electric field $\vec{E}$ exists and has a definite value at every point in space surrounding a source charge. If a field line had a break, it would imply that the electric field suddenly vanishes at that point, which is inconsistent with the continuous nature of the electrostatic force. Therefore, the field lines must be continuous curves.

(b) Field Lines Never Cross

If two electric field lines were to cross at a point $P$, it would mean that a test charge placed at $P$ would experience a force (and thus the electric field $\vec{E}$) tangent to both lines. This implies that the electric field at point $P$ has two different directions simultaneously. Since the electric field is a vector quantity, it can only have one unique direction at any given point. Hence, two field lines never cross.

1.8 Two point charges $q_A = 3\text{ $\mu$C}$ and $q_B = -3\text{ $\mu$C}$ are located $20\text{ cm}$ 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 $1.5 \times 10^{-9}\text{ C}$ is placed at this point, what is the force experienced by the test charge?

• $q_A = 3 \times 10^{-6}\text{ C}$, $q_B = -3 \times 10^{-6}\text{ C}$
• $d = 20\text{ cm}$, midpoint distance $r = 10\text{ cm} = 0.10\text{ m}$.

(a) Electric field at the midpoint $O$

The net field $\vec{E}$ is $\vec{E}_A + \vec{E}_B$.

• $\vec{E}_A$ (due to positive $q_A$) is directed away from $A$ (towards $B$).
• $\vec{E}_B$ (due to negative $q_B$) is directed towards $B$.The fields are in the same direction, so they add up. Since $|q_A| = |q_B|$ and $r$ is equal: $E_A = E_B$.$$E_A = k \frac{|q|}{r^2} = (9 \times 10^9) \frac{3 \times 10^{-6}}{(0.10)^2} = 2.7 \times 10^6\text{ N/C}$$$$E_{\text{net}} = E_A + E_B = 2 \times (2.7 \times 10^6\text{ N/C}) = 5.4 \times 10^6\text{ N/C}$$

The electric field at $O$ is $5.4 \times 10^6\text{ N/C}$ (directed from $A$ to $B$).

(b) Force experienced by the test charge

• Test charge $q_0 = -1.5 \times 10^{-9}\text{ C}$.
• $E_{\text{net}} = 5.4 \times 10^6\text{ N/C}$ (direction $A \to B$).

The force is $\vec{F} = q_0 \vec{E}$. Since $q_0$ is negative, $\vec{F}$ is opposite to $\vec{E}$.

$$F = |q_0| E_{\text{net}} = (1.5 \times 10^{-9}\text{ C}) (5.4 \times 10^6\text{ N/C})$$

$$F = 8.1 \times 10^{-3}\text{ N}$$

The force is $8.1 \times 10^{-3}\text{ N}$ (directed from $B$ to $A$).

1.9 A system has two charges $q_A = 2.5 \times 10^{-7}\text{ C}$ and $q_B = -2.5 \times 10^{-7}\text{ C}$ located at points $A: (0, 0, -15\text{ cm})$ and $B: (0, 0, +15\text{ cm})$, respectively. What are the total charge and electric dipole moment of the system?

Total Charge

$$Q_{\text{total}} = q_A + q_B = (2.5 \times 10^{-7}\text{ C}) + (-2.5 \times 10^{-7}\text{ C}) = 0$$

The total charge is zero.

Electric Dipole Moment ($\vec{p}$)

The magnitude is $p = q (2a)$, where $2a$ is the distance between the charges.

• $q = 2.5 \times 10^{-7}\text{ C}$
• $2a = (+15\text{ cm}) – (-15\text{ cm}) = 30\text{ cm} = 0.30\text{ m}$$$p = (2.5 \times 10^{-7}\text{ C}) (0.30\text{ m}) = 7.5 \times 10^{-8}\text{ C m}$$The vector $\vec{p}$ points from the negative charge ($q_B$ at $+15\text{ cm}$) to the positive charge ($q_A$ at $-15\text{ cm}$). This direction is along the negative $z$-axis.

The electric dipole moment has a magnitude of $7.5 \times 10^{-8}\text{ C m}$.

1.10 An electric dipole with dipole moment $4 \times 10^{-9}\text{ C m}$ is aligned at $30^\circ$ with the direction of a uniform electric field of magnitude $5 \times 10^4\text{ N C}^{-1}$. Calculate the magnitude of the torque acting on the dipole.

The magnitude of the torque $\tau$ is given by:

$$\tau = p E \sin\theta$$

• $p = 4 \times 10^{-9}\text{ C m}$
• $E = 5 \times 10^4\text{ N C}^{-1}$
• $\theta = 30^\circ$

$$\tau = (4 \times 10^{-9}) (5 \times 10^4) \sin(30^\circ)\text{ N m}$$

$$\tau = (20 \times 10^{-5}) (0.5)\text{ N m}$$

$$\tau = 10 \times 10^{-5}\text{ N m} = 1.0 \times 10^{-4}\text{ N m}$$

The magnitude of the torque is $1.0 \times 10^{-4}\text{ N m}$.

1.11 A polythene piece rubbed with wool is found to have a negative charge of $3 \times 10^{-7}\text{ C}$. (a) Estimate the number of electrons transferred (from which to which?) (b) Is there a transfer of mass from wool to polythene?

(a) Estimate the number of electrons transferred

Since charge is quantised, $q = n e$.

$$n = \frac{|q|}{e} = \frac{3 \times 10^{-7}\text{ C}}{1.602 \times 10^{-19}\text{ C}} \approx 1.87 \times 10^{12}$$

Since the polythene piece acquired a negative charge, it must have gained electrons. The wool, conversely, lost them.

Number of electrons transferred: $1.87 \times 10^{12}$.

Transfer is from wool to polythene.

(b) Is there a transfer of mass from wool to polythene?

Yes. Since electrons have mass ($m_e \approx 9.11 \times 10^{-31}\text{ kg}$), the transfer of electrons from wool to polythene results in a transfer of mass.

$$\text{Mass transfer} = n \times m_e \approx (1.87 \times 10^{12}) \times (9.11 \times 10^{-31}\text{ kg}) \approx 1.7 \times 10^{-18}\text{ kg}$$

There is a transfer of mass from wool to polythene, though it is negligibly small.

1.12 (a) Two insulated charged copper spheres A and B have their centres separated by a distance of $50\text{ cm}$. What is the mutual force of electrostatic repulsion if the charge on each is $6.5 \times 10^{-7}\text{ C}$? (b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

• $q = 6.5 \times 10^{-7}\text{ C}$, $r_1 = 50\text{ cm} = 0.50\text{ m}$

(a) Mutual force of electrostatic repulsion

$$F_1 = k \frac{q^2}{r_1^2} = (9 \times 10^9) \frac{(6.5 \times 10^{-7})^2}{(0.50)^2}\text{ N}$$

$$F_1 = (9 \times 10^9) \frac{42.25 \times 10^{-14}}{0.25}\text{ N} \approx 1.52 \times 10^{-2}\text{ N}$$

The force is $1.52 \times 10^{-2}\text{ N}$.

(b) Force of repulsion with doubled charge and halved distance

New charge $q’ = 2q$. New distance $r_2 = r_1/2$.

$$F_2 = k \frac{(q’)^2}{(r_2)^2} = k \frac{(2q)^2}{(r_1/2)^2} = k \frac{4q^2}{r_1^2/4} = 16 \left( k \frac{q^2}{r_1^2} \right)$$

$$F_2 = 16 F_1$$

$$F_2 = 16 \times (1.52 \times 10^{-2}\text{ N}) \approx 0.243\text{ N}$$

The new force of repulsion is $0.243\text{ N}$.

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 highest charge to mass ratio?

Assume the electric field $\vec{E}$ is directed from the bottom plate (positive) to the top plate (negative), or vice-versa. Let’s assume the top plate is positive and the bottom plate is negative, so $\vec{E}$ is pointing downward.

1. Particle 1: Deflects upward (towards the negative plate). It must be positively charged.
2. Particle 2: Travels undeflected. It must be uncharged (neutral).
3. Particle 3: Deflects downward (towards the positive plate). It must be negatively charged.

The vertical deflection ($y$) of a charged particle in a uniform field is proportional to the acceleration $a$, which is proportional to the charge-to-mass ratio ($q/m$): $y \propto a \propto q/m$.

Particle 3 shows the largest vertical deflection magnitude.

Therefore, Particle 3 has the highest charge to mass ratio $(|q/m|)$.

1.14 Consider a uniform electric field $\vec{E} = 3 \times 10^3 \hat{i}\text{ N/C}$. (a) What is the flux of this field through a square of $10\text{ cm}$ 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 $60^\circ$ angle with the $x$-axis?

• $E = 3 \times 10^3\text{ N/C}$ (in the $x$-direction).
• Area $A = (0.10\text{ m})^2 = 0.01\text{ m}^2$.

The electric flux is $\Phi = E A \cos\theta$.

(a) Plane is parallel to the $yz$ plane.

If the plane is parallel to the $yz$-plane, its area vector $\vec{A}$ is along the $x$-axis ($\hat{i}$). Since $\vec{E}$ is also along $\hat{i}$, the angle $\theta = 0^\circ$.

$$\Phi = E A \cos(0^\circ) = (3 \times 10^3) (0.01) (1) = 30\text{ N m}^2/\text{C}$$

The flux is $30\text{ N m}^2/\text{C}$.

(b) The normal to its plane makes a $60^\circ$ angle with the $x$-axis.

The angle $\theta = 60^\circ$.

$$\Phi = E A \cos(60^\circ) = (3 \times 10^3) (0.01) (0.5)$$

$$\Phi = 30 \times 0.5 = 15\text{ N m}^2/\text{C}$$

The flux is $15\text{ N m}^2/\text{C}$.

1.15 What is the net flux of the uniform electric field of Exercise 1.14 through a cube of side $20\text{ cm}$ oriented so that its faces are parallel to the coordinate planes?

• Uniform electric field $\vec{E} = 3 \times 10^3 \hat{i}\text{ N/C}$.
• Closed surface is a cube.

For a closed surface placed in a uniform electric field, the flux entering the surface (negative flux) is exactly equal to the flux leaving the surface (positive flux). The net enclosed charge is zero.

• Flux through the face perpendicular to the $x$-axis where the field enters ($\hat{A} = -\hat{i}$) is $\Phi_{\text{in}} = -E A$.
• Flux through the opposite face where the field leaves ($\hat{A} = +\hat{i}$) is $\Phi_{\text{out}} = +E A$.
• Flux through the other four faces (normal is $\pm\hat{j}$ or $\pm\hat{k}$) is $\Phi = E A \cos(90^\circ) = 0$.

The net flux $\Phi_{\text{net}} = \Phi_{\text{in}} + \Phi_{\text{out}} + 4\Phi_{90} = -E A + E A + 0 = 0$.

The net flux through the cube is zero.

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 $8.0 \times 10^3\text{ N m}^2/\text{C}$. (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?

(a) What is the net charge inside the box?

Use Gauss’s Law: $\Phi = Q_{\text{enclosed}} / \epsilon_0$.

$$Q_{\text{enclosed}} = \Phi \epsilon_0$$

$$Q_{\text{enclosed}} = (8.0 \times 10^3\text{ N m}^2/\text{C}) \times (8.854 \times 10^{-12}\text{ C}^2/\text{N m}^2)$$

$$Q_{\text{enclosed}} \approx 70.83 \times 10^{-9}\text{ C} \approx 7.08 \times 10^{-8}\text{ C}$$

The net charge inside the box is $7.08 \times 10^{-8}\text{ C}$.

(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?

If $\Phi_{\text{net}} = 0$, then $Q_{\text{enclosed}} = 0$. This means the net charge inside the box is zero.

Conclusion: No, you cannot conclude that there were no charges inside.

Reason: The box could contain equal and opposite charges (e.g., an electric dipole or multiple pairs of positive and negative charges) whose sum is zero, resulting in zero net flux.

1.17 A point charge $+10\text{ $\mu$C}$ is a distance $5\text{ cm}$ directly above the centre of a square of side $10\text{ cm}$. What is the magnitude of the electric flux through the square?

• $q = 10\text{ $\mu$C} = 10 \times 10^{-6}\text{ C}$.
• The charge is $5\text{ cm}$ above the centre of a $10\text{ cm}$ square.

The problem suggests viewing the square as one face of a $10\text{ cm}$ cube. Since the charge is $5\text{ cm}$ above the centre of this face, it is precisely at the centre of the imaginary cube.

By symmetry, the total flux $\Phi_{\text{total}}$ through the cube (which encloses the charge) is divided equally among the six faces.

$$\Phi_{\text{total}} = \frac{q}{\epsilon_0}$$

$$\Phi_{\text{square}} = \frac{1}{6} \Phi_{\text{total}} = \frac{q}{6 \epsilon_0}$$

$$\Phi_{\text{square}} = \frac{10 \times 10^{-6}\text{ C}}{6 \times (8.854 \times 10^{-12}\text{ C}^2/\text{N m}^2)}$$

$$\Phi_{\text{square}} \approx 1.88 \times 10^5\text{ N m}^2/\text{C}$$

The magnitude of the electric flux through the square is $1.88 \times 10^5\text{ N m}^2/\text{C}$.

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

• $Q_{\text{enclosed}} = 2.0\text{ $\mu$C} = 2.0 \times 10^{-6}\text{ C}$.
• Gaussian surface is a cube.

By Gauss’s Law, the net flux is independent of the shape or size of the Gaussian surface, as long as it encloses the charge.

$$\Phi_{\text{net}} = \frac{Q_{\text{enclosed}}}{\epsilon_0}$$

$$\Phi_{\text{net}} = \frac{2.0 \times 10^{-6}\text{ C}}{8.854 \times 10^{-12}\text{ C}^2/\text{N m}^2}$$

$$\Phi_{\text{net}} \approx 2.26 \times 10^5\text{ N m}^2/\text{C}$$

The net electric flux through the surface is $2.26 \times 10^5\text{ N m}^2/\text{C}$.

1.19 A point charge causes an electric flux of $-1.0 \times 10^3\text{ N m}^2/\text{C}$ to pass through a spherical Gaussian surface of $10.0\text{ cm}$ 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?

(a) Effect of Doubling the Radius

According to Gauss’s Law, the net electric flux through a closed surface depends only on the net charge enclosed, $Q_{\text{enclosed}}$. Changing the radius of the Gaussian surface does not change $Q_{\text{enclosed}}$.

The flux would remain $-1.0 \times 10^3\text{ N m}^2/\text{C}$.

(b) Value of the point charge

$$Q = \Phi \epsilon_0$$

$$Q = (-1.0 \times 10^3\text{ N m}^2/\text{C}) \times (8.854 \times 10^{-12}\text{ C}^2/\text{N m}^2)$$

$$Q = -8.854 \times 10^{-9}\text{ C}$$

The value of the point charge is $-8.85\text{ nC}$.

1.20 A conducting sphere of radius $10\text{ cm}$ has an unknown charge. If the electric field $20\text{ cm}$ from the centre of the sphere is $1.5 \times 10^3\text{ N/C}$ and points radially inward, what is the net charge on the sphere?

• $E = 1.5 \times 10^3\text{ N/C}$
• Distance $r = 20\text{ cm} = 0.20\text{ m}$

For a point outside a conducting sphere, the field is:

$$E = k \frac{|Q|}{r^2} \implies |Q| = \frac{E r^2}{k}$$

$$|Q| = \frac{(1.5 \times 10^3\text{ N/C}) (0.20\text{ m})^2}{9 \times 10^9\text{ N m}^2\text{ C}^{-2}}$$

$$|Q| = \frac{1.5 \times 10^3 \times 0.04}{9 \times 10^9}\text{ C} \approx 6.67 \times 10^{-9}\text{ C}$$

Since the electric field points radially inward, the source charge $Q$ must be negative.

The net charge on the sphere is $-6.67 \times 10^{-9}\text{ C}$.

1.21 A uniformly charged conducting sphere of $2.4\text{ m}$ diameter has a surface charge density of $80.0\text{ $\mu$C/m}^2$. (a) Find the charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere?

• Radius $R = 2.4\text{ m} / 2 = 1.2\text{ m}$.
• $\sigma = 80.0\text{ $\mu$C/m}^2 = 80.0 \times 10^{-6}\text{ C/m}^2$.

(a) Find the charge on the sphere

$$Q = \sigma A = \sigma (4 \pi R^2)$$

$$Q = (80.0 \times 10^{-6}\text{ C/m}^2) \times 4 \pi (1.2\text{ m})^2$$

$$Q = (80.0 \times 10^{-6}) \times 4 \pi (1.44)\text{ C} \approx 1.45 \times 10^{-3}\text{ C}$$

The charge on the sphere is $1.45 \times 10^{-3}\text{ C}$.

(b) What is the total electric flux leaving the surface of the sphere?

$$\Phi = \frac{Q}{\epsilon_0}$$

$$\Phi = \frac{1.45 \times 10^{-3}\text{ C}}{8.854 \times 10^{-12}\text{ C}^2/\text{N m}^2} \approx 1.64 \times 10^8\text{ N m}^2/\text{C}$$

The total electric flux is $1.64 \times 10^8\text{ N m}^2/\text{C}$.

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

The electric field $E$ for an infinite line charge with linear charge density $\lambda$ at a distance $r$ is:

$$E = \frac{\lambda}{2 \pi \epsilon_0 r} = 2 k \frac{\lambda}{r}$$

Solving for $\lambda$:

$$\lambda = \frac{E r}{2 k}$$

$$\lambda = \frac{(9 \times 10^4\text{ N/C}) (0.02\text{ m})}{2 \times (9 \times 10^9\text{ N m}^2\text{ C}^{-2})}$$

$$\lambda = \frac{18 \times 10^2}{18 \times 10^9}\text{ C/m} = 1 \times 10^{-7}\text{ C/m}$$

The linear charge density is $1.0 \times 10^{-7}\text{ C/m}$.

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 signs and of magnitude $17.0 \times 10^{-22}\text{ C/m}^2$. 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?

Let $\sigma = 17.0 \times 10^{-22}\text{ C/m}^2$. Plate 1 is positive ($+\sigma$), Plate 2 is negative ($-\sigma$).

The field due to a single infinite plate is $E_{\text{plate}} = \sigma / (2 \epsilon_0)$.

(a) In the outer region of the first plate

The fields from Plate 1 ($E_1$, away from the plate) and Plate 2 ($E_2$, towards the plate) are equal in magnitude and point in opposite directions.

$$E_{\text{outer}} = E_1 – E_2 = 0$$

The electric field is zero.

(b) In the outer region of the second plate

Similar to (a), the fields are equal in magnitude and point in opposite directions.

$$E_{\text{outer}} = E_1 – E_2 = 0$$

The electric field is zero.

(c) Between the plates

The field from Plate 1 ($E_1$) points from Plate 1 to Plate 2. The field from Plate 2 ($E_2$) points towards Plate 2 (i.e., from Plate 1 to Plate 2). Both fields are in the same direction and add up.

$$E_{\text{between}} = E_1 + E_2 = \frac{\sigma}{2 \epsilon_0} + \frac{\sigma}{2 \epsilon_0} = \frac{\sigma}{\epsilon_0}$$

$$E_{\text{between}} = \frac{17.0 \times 10^{-22}\text{ C/m}^2}{8.854 \times 10^{-12}\text{ C}^2/\text{N m}^2}$$

$$E_{\text{between}} \approx 1.92 \times 10^{-10}\text{ N/C}$$

The electric field between the plates is $1.92 \times 10^{-10}\text{ N/C}$ (directed from the positive plate to the negative plate).