Electric Charges And Fields

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Electric Charges and Fields – Exercises

1.1 What is the force between two small charged spheres having charges of

{2 × 10^{-7}~C}

and

{3 × 10^{-7}~C}

placed

{30~cm}

apart in air?

Given that

{q_1}

{= 2 × 10^{-7}~C}

{q_2}

{= 3 × 10^{-7}~C}

{r}

{= 30~cm}

{= 0.3~m}

We know that, as per Coulomb’s Law:

{F = \dfrac{1}{4πϵ_0} \cdot \dfrac{q_1 q_2}{r^2}}

Substituting the values, we get

{F}

{= 9 × 10^9 × \dfrac{(2 × 10^{-7}) × (3 × 10^{-7})}{(0.3)^2}}

{= 9 × 10^9 × \dfrac{(2 × 10^{-7}) × (3 × 10^{-7})}{(3 × 10^{-1})^2}}

{= 9 × 10^9 × \dfrac{(2 × 10^{-7}) × (3 × 10^{-7})}{9 × 10^{-2}}}

{= 6 × 10^{-3}~N}

∴ The force between the given two small charged spheres is

{6 × 10^{-3}~N}

⚡Note: As the resulting force is positive and also as the two given charges are like charges, we can say that the force between the given two charges is repulsive in nature.

1.2 The electrostatic force on a small sphere of charge

{0.4~μC}

due to another small sphere of charge

{-0.8~μC}

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?

(a) To find the distance between the two spheres:

Given that

{q_1}

{= 0.4~μC}

{q_2}

{= -0.8~μC}

{F}

{= 0.2~N}

{r}

= ?

We know that, as per Coulomb’s Law:

{F = \dfrac{1}{4\pi\varepsilon_0} × \dfrac{q_1 q_2}{r^2}}

Where, ϵ0 is the permittivity of the free space and is given by

{\dfrac{1}{4πϵ_0} = 9 × 10^9 Nm^2C^{-2}}

Rearranging, we get

{r^2 = \dfrac{1}{4\pi\varepsilon_0} × \dfrac{q_1 q_2}{F}}

Substituting the values, we get

{r^2}

{= 9 × 10^9 × \dfrac{(0.4 × 10^{-6})(0.8 × 10^{-6})}{0.2}}

{= 144 × 10^{-4}}

{∴ r}

{= \sqrt{144 × 10^{-4}}}

= 12 × 10-2

{= 0.12~m}

∴ The distance between the two spheres is

{0.12~m}

(b) Force on the second sphere due to the first:

The attractive force exerted by each sphere on the other is equal in magnitude.

∴ The force on the second sphere due to the first is

{0.2~N}

1.3 Check that the ratio

{\dfrac{ke^2}{Gm_em_p}}

is dimensionless. Look up a table of physical constants and determine the value of this ratio. What does the ratio signify?

The various constants in

{\dfrac{ke^2}{Gm_em_p}}

, their values and dimensions are as follows:

Constant

Definition

(Approximate)
Numeric
Value

Unit

{k}

Coloumb’s Constant

{9 × 10^9}

{Nm^2 / C^2}

{e}

Charge of electron

{1.6 × 10^{-19}}

{G}

Gravitational Constant

{6.67 × 10^{-11}}

{Nm^2/kg^2}

{m_e}

Mass of electron

{9.11 × 10^{-31}}

{kg}

{m_p}

Mass of proton

{1.67 × 10^{-27}}

{kg}

To Check whether the ratio

{\dfrac{ke^2}{Gm_em_p}}

is dimensionless

Substituting the above units into the ratio

{\dfrac{ke^2}{Gm_em_p}}

, we have

{\dfrac{ke^2}{Gm_em_p}}

{= \dfrac{\left(\dfrac{Nm^2}{C^2}\right).C^2}{\left(\dfrac{Nm^2}{kg^2}\right) \cdot kg \cdot kg}}

{= \dfrac{\left(\dfrac{\cancel{Nm^2}}{\cancel{C^2}}\right).\cancel{C^2}}{\left(\dfrac{\cancel{Nm^2}}{\cancel{kg^2}}\right) \cdot \cancel{kg} \cdot \cancel{kg}}}

{= M^0L^0T^0}

Thus we can see that the ratio

{\dfrac{ke^2}{Gm_em_p}}

is dimensionless

To Determine the value of the ratio

{\dfrac{ke^2}{Gm_em_p}}

Substituting the values of various constants from the above given table into the ratio

{\dfrac{ke^2}{Gm_em_p}}

, we have,

{\dfrac{ke^2}{Gm_em_p}}

{= \dfrac{(9 × 10^9) × (1.6 × 10^{-19})^2}{(6.67 × 10^{-11}) × (9.11 × 10^{-31}) × (1.67 × 10^{-27})}}

{\approx 2.3 × 10^{39}}

The value of the ratio is approximately

{ 2.3 × 10^{39}}

This shows that the electric force is about

{10^{39}}

times stronger than gravitational force.

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?

The statement “electric charge of a body is quantized” means that a body can possess charge only in discrete amounts—specifically, as integral (1, 2, 3, …..

{n}

) multiples of the elementary charge

{e}

. This implies that the transfer of charge occurs in whole numbers of electrons (or protons), never in fractions. Mathematically, total charge

{q}

on a body is given by

{q=\pmne}

where

●

{n}

is an integer and

●

{e = 1.6 × 10^{-19} C}

Thus, charge is not continuous but comes in fixed packets. In otherwords, we can say that the ‘electric charge of a body is quantised’

On the macroscopic scale, however, the amount of charge involved is extremely large compared to the elementary charge. In such cases, the discrete nature of charge becomes negligible, and the variation in charge appears smooth and continuous. Therefore, for practical purposes in large-scale systems, the quantization of electric charge is ignored, and charge is treated as a continuous variable.

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.

When a glass rod is rubbed with a silk cloth, charges of equal magnitude but opposite nature appear on both bodies—glass becomes positively charged, and silk becomes negatively charged. This occurs because electrons are transferred from the glass rod to the silk cloth. Importantly, no new charge is created in the process; charge is only redistributed between the two bodies. This is known as charging by friction, where charges are produced in equal and opposite pairs. The total charge of the system before and after rubbing remains zero, which confirms that the phenomenon is consistent with the law of conservation of charge. Similar observations are made with many other pairs of materials, highlighting that the net charge in an isolated system always remains constant.

1.6 Four point charges

{q_A = 2~μC}

{q_B = –5~μC}

{q_C = 2~μC}

, and

{q_D = –5~μC}

are located at the corners of a square ABCD of side

{10~cm}

. What is the force on a charge of

{1~μC}

placed at the centre of the square?

The charges at the points A and C are equal in magnitude. So, the force exerted by the charges at A and C on O will be equal repulsive force acting in opposite direction. Hence they’ll cancel each other.

Similarly, the charges at the points B and D are equal in magnitude. So, the force exerted by the charges at B and D on O will be equal attractive force acting in opposite direction. Hence they’ll cancel each other.

Hence, the resulting force on the charge placed at the center of the square O will be 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)

An electrostatic field line is a continuous curve because it represents the path traced by a unit positive charge placed in the field, which always experiences a continuous and smooth force. Since the electric field varies smoothly in space, a test charge does not jump between positions. Hence, field lines cannot have sudden breaks or discontinuities.

(b)

Two field lines can never intersect because if they did, the electric field at that point would have two different directions simultaneously. This contradicts the fundamental property of electric fields, which have a unique direction at every point in space. Therefore, field lines never cross each other.

1.8 Two point charges

{q_A = 3~μC}

and

{q_B = –3~μC}

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?

a. To find the electric field at the mid-point

{O}

of the two points

{A}

and

{B}

Given that

{q_A}

{= 3 × 10^{-6}~C}

{q_B}

{= -3 × 10^{-6}~C}

{AB}

{= 20~cm}

{O}

is the mid-point of

{A}

and

{B}

, we have

{OA = OB}

{= 10~cm}

{= 0.1~m}

Now, the electric field at

{O}

due to the charge at

{A}

is given as,

{E_{AO} = \dfrac{1}{4πϵ_0} . \dfrac{q_A}{r^2}}

along AO (which is same as along OB)

And the electric field at

{O}

due to the charge at

{B}

is given as,

{E_{OB} = \dfrac{1}{4πϵ_0} \cdot \dfrac{q_B}{r^2}}

along OB

Where

{ϵ_0}

is the permittivity of free space.

We also know that

{\dfrac{1}{4πϵ_0} = 9 × 10^9~Nm^2/C^2}

Substituting these values, we get

{E_{AO}}

{= E_{OB}}

{= \left|\dfrac{1}{4πϵ_0} × \dfrac{q_B}{r^2}\right|}

along OB

{= \left|9 × 10^9 × \dfrac{3 × 10^{-6}}{(0.1)^2}\right|}

along OB

{= 2.7 × 10^6~N/C}

along OB

∴ The total electric field

{E}

at point

{O}

is given as

{E}

{= E_{AO} + E_{OB}}

{= (2.7 × 10^6) + (2.7 × 10^6)}

{= 5.4 × 10^6~N/C}

along OB

∴ The total electric field at the mid-point

{O}

of the points

{A}

and

{B}

{5.4 × 10^6~N/C}

along OB

(b) Force experienced by the negative test charge when it is placed at the point

{O}

The charge placed at point

{O}

{q_o = -1.5 × 10^{-9}~C}

Now the force

{F}

experienced by this charge

{q_o}

when placed in an electric field

{E}

is given as

{F}

{= q_oE}

{= (-1.5 × 10^{-9}) . (5.4 × 10^6)}

{= -8.1 × 10^{-3}~N}

The negative sign indicates that the direction of this force is opposite to that of electric field i.e., in the direction of

{OA}

So, the force on this negative test charge is equal to

{8.1 × 10^{-3}~N}

acting along

{OA}

1.9 A system has two charges

{q_A = 2.5 × 10^{–7}~C}

and

{q_B = –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?

Given that

Charge at A,

{q_A}

{= 2.5 × 10^{-7}~C}

Charge at B,

{q_B}

{= -2.5 × 10^{-7}~C}

As both the points

{A (0, 0, -15)}

and

{B (0, 0, 15)}

are on the z-axis, the distrance between the two points

{d}

is equal to the sum of their distances from the orgin

{O (0, 0, 0)}

{d}

{= OA + OB}

{= 15~cm + 15~cm}

{= 30~cm}

{= 0.3~m}

To find the total charge:

The total charge of the given system will be the sum of the individual charges.

Total charge

{q}

{= q_A + q_B}

{= (2.5 × 10^{-7}~C) + (-2.5 × 10^{-7}~C)}

= 0

∴ The total charge is

{0~C}

To find the electric dipole moment of the system:

The electric dipole moment is defined as the product of the charge at any of the given points and the distance between them.

{\vec{p}}

{= \text{electric charge} × AB}

{= qd}

{= 2.5 × 10^{-7} × 0.3}

{= 7.5 × 10^{-8}~Cm}

∴ The electric dipole moment is

{7.5 × 10^{-8}~Cm}

and the direction is from

{B}

{A}

along

{z\text{-axis}}

⚡Note: The direction of the electric dipole moment is from negative charge to the positive charge. Also note that the unit of dipole momentum is Coulomb-meter (not to be confused as centimeter)

1.10. An electric dipole with dipole moment

{4 × 10^{–9}~Cm}

is aligned at 30° with the direction of a uniform electric field of magnitude

{5 × 10^4 NC^{–1}}

.
Calculate the magnitude of the torque acting on the dipole.

Given that

Electric dipole moment,

{\vec{p} = 4 × 10^{-9}~Cm}

Electric Field,

{E = 5 × 10^4~NC^{-1}}

Angle between the electric dipole moment

{\vec{p}}

and the electric field

{E}

is given as

{θ = 30°}

The magnitude of the torque

{τ}

acting on the dipole is given as

{τ}

{= \vec{p}E \sin θ}

{= 4 × 10^{-9} × 10^4 \sin 30°}

{= 20 × 10^{-5} × \dfrac{1}{2}}

{= 10^{-4}~Nm}

∴ The magniture of the torque

{τ}

acting on the dipole is

{10^{-4}~Nm}

1.11

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?

Rubbing polythene with wool results in the transfer of electrons from the wool to the polythene. This electron transfer causes the polythene to become negatively charged and the wool to become positively charged.

a. To estimate the number of electrons transferred:

Given that:

Charge acquired by polythene,

{q = -3 × 10^{-7}~C}

We know that

Elementary charge (charge of one electron),

{e = -1.6 × 10^{-19}}

We know that the relation between the total charge

{q}

to the number of electrons transferred

{n}

and the elementary charge of electron

{e}

is given as

{q = ne}

Rearranging to calculate the number of electrons, we get

{n}

{= \dfrac{q}{e}}

{= \dfrac{-3 × 10^{-7}}{-1.6 × 10^{-19}}}

{≈ 1.875 × 10^{12}}

∴ Approximately,

{1.875 × 10^{12}}

electrons are transferred from the wool to the polythene.

(b). To check whether there is a transfer of mass from wool to polythene:

We know that the electrons have their own mass and hence when the electrons are transferred from one object to another object it results in a corresponding mass transfer too. Hence, there is definitely a mass transfer from wool to polythene.

We know that the mass of one electron is

{m_e = 9.1 × 10^{-31}~kg}

And we have just now calculated above that the number of electrons transferred are

{n ≈ 1.875 × 10^{12}}

So, the total mass transferred from wool to polythene can be computed as

{m}

{= n × m_e}

{= (1.875 × 10^{12}) × (9.1 × 10^{-31})}

{= 1.706 × 10^{-18}~kg}

∴ Approximately

{1.706 × 10^{-18}~kg}

of mass of electrons is transferred from wool to polythene.

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 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?

(a) To calculate the mutual force of electrostatic repulsion when the charge on each sphere is

{6.5 × 10^{-7}~C}

The charges on the spheres

{A}

and

{B}

are given as:

{q_A}

{= 6.5 × 10^{-7}~C}

{q_B}

{= 6.5 × 10^{-7}~C}

And the distance

{r}

between the two spheres is given as

{r}

{= 50~cm}

{= 0.5~m}

As both the charges are positive, there will be repulsive force between them which can be calculated as

{F = \dfrac{1}{4πϵ_0} . \dfrac{q_Aq_B}{r^2}}

Where

{\dfrac{1}{4πϵ_0}}

is permittivity of free-space and is given as

{\dfrac{1}{4πϵ_0} = 9 × 10^9~Nm^2C^{-2}}

Substituting, we get the force of repulsion between the given charges as

{F}

{= \dfrac{1}{4πϵ_0} . \dfrac{q_Aq_B}{r^2}}

{= 9 × 10^9 × \dfrac{6.5 × 10^{-7} × 6.5 × 10^{-7}}{(0.5)^2}}

{= 1.52 × 10^{-2}~N}

∴ The mutual force of electrostatic repulsion between the two charges

{A}

and

{B}

in this case is

{1.52 × 10^{-2}~N}

(b) To calculate the force of repulsion when the charge on each sphere is doubled and the distance is halved.

In this case, the charges on the spheres will be,

{q_A = q_B}

{= 2 × 6.5 × 10^{-7}}

{= 1.3 × 10^{-6} C}

And the halved distance between the spheres will be

{r}

{= \dfrac{0.5}{2}}

{= 0.25~m}

Now, the force of replusion between the two spheres can be calculated as

{F}

{= \dfrac{1}{4πϵ_0} × \dfrac{q_Aq_B}{r^2}}

{= 9 × 10^9 × \dfrac{(1.3 × 10^{-6}) × (1.3 × 10^{-6})}{(0.25)^2}}

{= 16 × 1.52 × 10^{-2}}

{= 0.243~N}

∴ The mutual force of repulsion when the charge on each sphere is doubled and the distance beween them is halved is

{0.243~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?

Figure 1.30

To Determine the Signs of the Charges:

In a uniform electrostatic field, charged particles experience a force that causes them to deflect. The direction of this deflection depends on the sign of the charge:

●

Positively charged particles are attracted towards the negative plate.

●

Negatively charged particles are attracted towards the positive plate.

Observing the trajectories:

●

Particles 1 and 2 are deflected towards the positive plate, indicating they are negatively charged.

●

Particle 3 is deflected towards the negative plate, indicating it is positively charged.

To Identify the Particle with the Highest Charge-to-Mass Ratio:

The deflection

{y}

of a charged particle in an electric field is given by the equation:

{y = \dfrac{1}{2} . \dfrac{qE}{m} . t^2}

Where:

●

{q}

is the charge,

●

{E}

is the electric field strength,

●

{m}

is the mass,

●

{t}

is the time the particle spends in the field.

Assuming all particles enter the field with the same initial velocity and spend equal time

{t}

in the field, the deflection

{y}

is directly proportional to the charge-to-mass ratio

{\dfrac{q}{m}}

{\dfrac{q}{m} ∝ y}

From the figure, Particle 3 exhibits the maximum deflection, implying it has the highest charge-to-mass ratio among the three particles.

1.14

Consider a uniform electric field

{E = 3 × 10^3~\hat{i}~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?

(a) To find the Flux through a square parallel to the

{yz}

-plane:

Given that

Electric field,

{\vec{E}}

{= 3 × 10^3~\hat{i}~N/C}

Side of the square,

{l}

{= 10~cm}

{= 0.1~m}

Area of the square,

{A}

{= l^2}

{= (0.1)^2}

{= 0.01~m^2}

As the square lies in the

{yz}

-plane, its area vector

{\vec{A}}

is along the x-axis

{(\hat{i})}

Electric flux

{(Φ_E)}

is given by:

{Φ_E}

{= \vec{E} . \vec{A}}

{= EA\cos θ}

Where

{θ}

is the angle between

{\vec{E}}

and

{\vec{A}}

Since both

{\vec{E}}

and

{\vec{A}}

are along

{\hat{i}}

{\theta = 0°}

So,

{\cos θ = 1}

∴

{Φ_E}

{= 3 × 10^3 × 0.01 × \cos0°}

{= 30~Nm^2/C}

∴ The electric flux through the square is

{30~Nm^2/C}

(b) To find the Flux through the square inclined at 60° to the

{x}

-axis:

Now, the angle between

{\vec{E}}

and

{\vec{A}}

{θ = 60°}

∴

{Φ_E}

{= 3 × 10^3 × 0.01 × \cos60°}

{= 3 × 10^3 × 0.01 × \dfrac{1}{2}}

{= 15~Nm^2/C}

∴ The electric flux through the inclined square is

{15~Nm^2/C}

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 coordinate planes?

Given that

Uniform Electric Field

{\vec{E}}

{= 3 × 10^3~\hat{i}~N/c}

Side of the cube,

{l}

{= 20~cm}

{= 0.2~m}

It is also given that the faces of the cube are parallel to the co-ordinate plane.

Now, the electric flux through any surface is given by

{Φ_E = \vec{E} \cdot \vec{A}}

Where

{\vec{E}}

is the electric field vector,

{\vec{A}}

is the area vector perpendicular to the surface,

{θ}

is the angle between

{\vec{E}}

and

{\vec{A}}

Now, in the given scenario,

●

Two faces of the cube are perpendicular to the x-axis (i.e., normal vectors along

{\hat{i}}

and

{-\hat{i}}

●

The other four faces are parallel to the x-axis (i.e., their normal vectors are perpendicular to

{\vec{E}}

, i.e., normal vectors

{\hat{j}}

and normal vector

{-\hat{j}}

. Due to space constraint the other two normal vectors

{\hat{k}}

and

{-\hat{k}}

are not shown in the figure.).

For faces perpendicular to

{\vec{E}}

i.e, for faces whose normal vectors are

{\hat{i}}

and

{-\hat{i}}

For face with normal

{\hat{i},~Φ_1}

{= EA~.~\cos 0°}

{= EA~.~1}

{= EA}

For face with normal

{-\hat{i},~Φ_2}

{= EA~.~\cos 180°}

{= EA~.~(-1)}

{= -EA}

For faces parallel to

{\vec{E}}

i.e, for faces whose normal vectors are

{\hat{i}, -\hat{i}, \hat{k} \text{ and } -\hat{k}}

{Φ_p}

{= EA~.~\cos 90°}

{= EA~.~0}

{= 0}

∴ The net flux through (all six faces of) the cube is given as

{Φ}

{= (Φ_1) + (Φ_2) + (4 × Φ_p)}

{= EA + (-EA) + (4 \times 0)}

{= EA - EA + 0}

{= 0}

∴ The net flux through the cube is

{0~Nm^2/C}

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 × 10^3~Nm^2/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) To calculate the Net Charge inside the box:

To determine the net charge enclosed inside the box, we use Gauss’s Law, which relates the electric flux through a closed surface to the charge enclosed. So, as per Gauss’s Law, we have

{Φ_E = \dfrac{q_{net}}{ϵ_0}}

Where

–

{Φ_E}

is the electric flux,

–

{q_{net}}

is the net charge enclosed,

–

{ϵ_0}

is the permittivity of freespace, and is given as

{8.854 × 10^{-12}~C^2/Nm^2}

Now, it is given in the problem that

{Φ_E = 8.0 × 10^3~Nm^2/C}

Rearranging Gauss’s Law to solve for

{q_{net}}

, we get

{q_{net} = Φ_E \cdot ϵ_0}

Substituting the given and known values, we get

{q_{net}}

{= (8.0 × 10^3) × (8.854 × 10^{-12})~C}

{= 7.08 × 10^{-8}~C}

{= 0.0708 × 10^{-6}~C}

{= 0.0708~μC}

∴ The charge inside the box is approximately

{0.0708~μC}

(b) Interpreting the Net Outward Flux:

If the net outward electric flux through the surface of the box is zero, i.e.,

{Φ_E = 0}

, applying Gauss’s Law,

{Φ_E}

{= \dfrac{q_{net}}{ϵ_0}}

{⇒ q_{net}}

{= Φ_E \cdot ϵ_0}

= 0

This implies that the net charge enclosed within the box is zero.

However, this does not necessarily mean that there are no charges inside the box. It’s possible that there are equal amounts of positive and negative charges whose effects cancel each other out, resulting in a net charge of zero.

So, to answer the question, we can say that No, we canno conclude that there are no charges inside the box. A net flux of zero indicates that the total enclosed charge is zero, but there could still be positive and negative charges present in equal amounts that sum to zero.

1.17 A point charge

{+10~μC}

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}

Figure 1.31

To determine the electric flux through the square, we can utilize Gauss’s Law, which relates the electric flux through a closed surface to the charge enclosed:

{Φ_E = \dfrac{q_{net}}{ϵ_0}}

Where:

●

{Φ_E}

is the electric flux,

●

{q_{net}}

is the enclosed charge,

●

{ϵ_0 = 8.854 × 10^{-12}~C^2/Nm^2}

is the vacuum permittivity.

Step 1: Conceptualizing the Setup

Imagine constructing a cube with side length

{10~cm}

such that the square in question is one of its faces. Placing the

{+10~μC}

point charge at the center of this cube (which is

{5~cm}

above the square) ensures symmetry.

Step 2: Applying Gauss’s Law

Since the cube is symmetrical and the charge is at its center, the electric flux will be evenly distributed through all six faces of the cube. Therefore, the flux through one face (our square) is:

{Φ_{\text{square}} = \dfrac{1}{6} \cdot \dfrac{q}{ϵ_0}}

Step 3: Substituting in the Values

Given that:

{q}

{ = 10~μC}

{= 10 × 10^{-6}~C}

Substituting the values of

{q}

and

{ϵ_0}

, we have

{Φ_\text{square}}

{= \dfrac{q_{net}}{ϵ_0}}

{= \dfrac{1}{6} \cdot \dfrac{10 × 10^{-6}}{8.854 × 10^{-12}}}

{≈ 1.88 × 10^5~Nm^2/C}

∴ The magnitude of the electric flux through the square is approximately

{1.88 × 10^5~Nm^2/C}

1.18 A point charge of

{2.0~μC}

is at the centre of a cubic Gaussian surface

{9.0~cm}

on edge. What is the net electric flux through the surface?

The enclosed electric charge is given as

{q_{net}}

{= 2.0~μC}

{= 2.0 × 10^{-6}~C}

To determine the net electric flux through the surface of the cube, we utilize Gauss’s Law, which relates the electric flux through a closed surface to the charge enclosed:

{Φ_E = \dfrac{q_{net}}{ϵ_0}}

Where,

●

{Φ_E}

is the electric flux,

●

{q_{net}}

is the enclosed charge,

●

{ϵ_0 = 8.854 × 10^{-12}~C^2/Nm^2}

is the permittivity of the free space.

{∴ Φ_E}

{= \dfrac{2.0 × 10^{-6}~C}{8.854 × 10^{-12}~C^2/Nm^2}}

{≈ 2.26 × 10^5~Nm^2/C}

∴ The net electric flux through the surface is approximately

{2.26 × 10^5~Nm^2/C}

1.19 A point charge causes an electric flux of

{–1.0 × 10^3~Nm^2/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?

(a) Effect of Doubling the Radius on Electric Flux:

According to Gauss’s Law, the electric flux

{Φ_E}

through a closed surface is given by:

{Φ_E = \dfrac{q_{net}}{ϵ_0}}

Where,

●

{Φ_E}

is the electric flux,

●

{q_{net}}

is the net charge eclosed within the surface,

●

{ϵ_0 = 8.854 × 10^{-12}~C^2/Nm^2}

is the permittivity of free space.

Key Insight: The electric flux through a closed surface depends solely on the enclosed charge, not on the size or shape of the surface.

∴ Even of the radius of the spherical Gaussian surface is doubleed, the enclosed charge remains unchanged, and thus, the electric flux remains the same

(b) Calculating the Value of the Point Charge:

The electric flux is given as

Electric flux,

{Φ_E = -1.0 × 10^3~Nm^2/C}

As per Gaussian Law:

{q_{net} = Φ_E × ϵ_0}

Substituting the known values,

{q_{net}}

{= (-1.0 × 10^3) × (8.854 × 10^{-12})~C}

{= -8.854 × 10^{-9}~C}

∴ The value of the point charge is approximately

{-8.85~nC}

(nano Coulombs)

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 × 10^3~N/C}

and points radially inward, what is the net charge on the sphere?

To determine the net charge on a conducting sphere given the electric field at a point outside the sphere, we can use the formula for the electric field due to a point charge (which also applies to uniformely charged conducting sphere):

{E = \dfrac{1}{4πϵ_0} \cdot \dfrac{q}{r^2}}

Where,

●

{E}

is the electric field magnitude,

●

{q}

is the net charge on the sphere

●

{r}

is the distance from the center of the sphere to the point where the field is measured,

●

{ϵ_0 = 8.854 × 10^{-12}~C^2/Nm^2}

is the permittivity of free space.

Also, it is given in the problem that

{E}

{= 1.5 × 10^3~N/C}

(directed radially inward)

{r}

{= 20~cm}

{= 0.20~m}

To Calculate the Net Charge on the Sphere:

Rewriting the formula

{E = \dfrac{1}{4πϵ_0} \cdot \dfrac{q}{r^2}}

to solve for

{q}

, we have

{q = E \cdot 4πϵ_0 \cdot r^2}

Substituting the known values, we get,

{q}

{= (1.5 × 10^3) \cdot 4π \cdot (8.854 × 10^{-12}) \cdot (0.20)^2}

{≈ 6.67 × 10^{-9}~C}

Since, the electric field is directed radially inward, the net charge on the sphere is negative.

{q}

{≈ -6.67 × 10^{-9}~C}

{= -6.67~nC}

∴ The net charge on the sphere is approximately

{-6.67~nC}

(nano Coulombs)

1.21 A uniformly charged conducting sphere of

{2.4~m}

diameter has a surface charge density of

{80.0~ μC/m^2}

. (a) Find the charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere?

Given that

Diameter of the sphere,

{d}

{= 2.4~m}

Radius of the sphere,

{r}

{= \dfrac{d}{2}}

{= 1.2~m}

Surface charge density,

{σ}

{= 80.0~μC/m^2}

{= 80.0 × 10^{-6}~C/m^2}

Vacuum permittivity,

{ϵ_0}

{= 8.854 × 10^{-12}~C^2/Nm^2}

(a) To calculate the total charge Q:

The total charge on the surface of the sphere is given by:

{Q = σ \times A}

Where

{A}

is the surface area of the sphere and is given as:

{A}

{= 4πr^2}

{= 4π(1.2)^2}

{= 4π \times 1.44}

{= 18.0956~m^2}

{∴~Q}

{= 80.0 × 10^{-6}~C/m^2 × 18.0956~m^2}

{= 1.4476 × 10^{-3}~C}

∴ The total charge on the sphere is approximately

{1.45 × 10^{-3}~C}

(b) Calculating the Total Electric Flux

{Φ}

According to Gauss’s Law, the total electric flux through a closed surface is:

{Φ}

{= \dfrac{Q}{ϵ_0}}

{= \dfrac{1.4476 × 10^{-3}~C}{8.854 × 10^{-12}~C^2/Nm^2}}

{≈ 1.634 × 10^8~Nm^2/C}

∴ The total electric flux leaving the surface of the sphere is approximately

{1.63 × 10^8~Nm^2/C}

1.22 An infinite line charge produces a field of

{9 × 10^4~N/C}

at a distance of

{2~cm}

. Calculate the linear charge density.

To determine the linear charge density (

{λ}

) of an infinite line charge that produces an electric field (

{E}

) of

{9 × 10^4~N/C}

at a distance of (

{r}

) of

{2~cm}

, we can use the formula for the elecric field due to an infinite line charge, which is given as,

{E = \dfrac{λ}{2πϵr}}

Where

●

{E}

is the electric field,

●

{λ}

is the linear charge density,

●

{ϵ_0 = 8.854 × 10^{-12}~C^2/Nm^2}

is the vacuum permittyvity (permittivity of free space),

●

{r = 2~cm = 0.02~m}

is the distance from the line charge

Re-arranging the formula to solve for

{λ}

, we have

{λ = 2πϵ_0rE}

Substituting the known/given values

{λ}

{= 2π × 8.854 × 10^{-12}~C^2/Nm^2 × 0.02~m × 9 × 10^4 N/C}

{≈ 1.0 × 10^{-7}~C/m}

∴ The linear charge density is approximately

{1.0 × 10^{-7}~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 × 10^{–22}~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?

Electric Field in different regions of parallel plates close to each other.

To determine the electric field E in various regions around two large, thin, parallel metal plates with surface charge densities of opposite signs and magnitude

{σ = 17.0 × 10^{-22}~C/m^2}

, we apply Gauss’s law.

(a)

{E}

in the outer region of the first plate:

In the region outside the first plate (say, to the left of plate A, shown by region I in the figure), the electric fields due to both plates are equal in magnitude but opposite in direction. Therefore, they cancel each other, resulting in a net electric field:

{E_{\text{outside}} = 0~N/C}

(b)

{E}

in the outer region of the second plate.

Similarly, in the region outside the second plate (to the right of plate B, shown by the region III in the figure), the electric fields from both plates cancel each other:

{E_{\text{outside}} = 0~N/C}

(c)

{E}

between the plates:

Between the plates, the electric fields due to both plates add up in the same direction. The electric field due to single infinite sheet is

{E = \dfrac{σ}{2ϵ_0}}

, so the total electric field between the plates is:

{E_\text{between} = \dfrac{σ}{ϵ_0}}

Substituting the values,

{E_\text{between}}

{= \dfrac{σ}{ϵ_0}}

{= \dfrac{17.0 × 10^{-22}}{8.854 × 10^{-22}}}

{≈ 1.92 × 10^{-10}~N/C}

To summarize

(a)

In the outer region of the first plate:

{E = 0~N/C}

(b)

In the outer region of the second plate:

{E = 0~N/C}

(c)

Between the plates:

{E ≈ 1.92 × 10^{-22}~N/C}