Class 12 Physics Ch Electric Charges and Fields

In our everyday life, we experience the effects of electric charges without always realizing it. From the simple shock we get when touching a doorknob after walking on a carpet to the complex working of electronic devices, electric charges play a fundamental role in shaping the physical world around us.

In the Class 12 Physics Ch Electric Charges and Fields, you will dive deep into the concept of electric charge—one of the most basic and crucial properties of matter. We will explore how charges interact, the behavior of static electricity, and the principles governing these interactions. This chapter sets the foundation for understanding electric forces, electric fields, and their importance in both natural phenomena and technological applications.

Key Concepts Covered: Class 12 Physics Ch Electric Charges and Fields

1. Electric Charge:
2. Coulomb’s Law:
3. Electric Field:
4. Electric Dipole:
5. Gauss’s Law:

Class 12 Physics Ch Electric Charges and Fields – Solutions

Additional Exercises
Question 1.25
An oil drop of 12 excess electrons is held stationary under a constant electric
field of 2.55 × 104 N C−1 in Millikan’s oil drop experiment. The density of the
oil is 1.26 g cm−3. Estimate the radius of the drop. (g = 9.81 m s−2; e = 1.60 ×
10−19 C).

Answer :

Given:
Number of excess electrons (n) = 12
Electric field (E) = 2.55 × 10⁴ N/C
Charge (q) = n × e = 12 × 1.60 × 10⁻¹⁹ = 1.92 × 10⁻¹⁸ C
Density of oil (ρ) = 1.26 × 10³ kg/m³
Gravitational acceleration (g) = 9.81 m/s²

When the drop is stationary:
Electric force = Gravitational force
So,
qE = mg

Mass of spherical drop:
m = (4/3) × π × r³ × ρ

Now,
qE = (4/3) × π × r³ × ρ × g

Solving for r³:
r³ = (3 × q × E) / (4 × π × ρ × g)

Substitute values:
r³ = (3 × 1.92 × 10⁻¹⁸ × 2.55 × 10⁴) / (4 × 3.1416 × 1.26 × 10³ × 9.81)
r³ ≈ 9.47 × 10⁻¹⁸ m³

Take cube root:
r ≈ (9.47 × 10⁻¹⁸)^(1/3)
r ≈ 2.08 × 10⁻⁶ m

Final Answer:
Radius of the oil drop ≈ 2.08 micrometres (μm)

Question 1.27
In a certain region of space, electric field is along the z-direction throughout.
The magnitude of electric field is, however, not constant but increases
uniformly along the positive z-direction, at the rate of 105 NC−1 per metre.
What are the force and torque experienced by a system having a total dipole
moment equal to 10−7 Cm in the negative z-direction?
Answer :

Given:

• Electric field direction: along the z-axis
• Electric field varies uniformly:
  dE/dz = 10⁵ N/C per metre
• Dipole moment (p) = 10⁻⁷ C·m, along the negative z-direction

Force on a Dipole in a Non-uniform Electric Field:

In a non-uniform electric field, the net force on a dipole is given by:

F = (p · ∇)E

Here, the dipole moment is along the z-direction, and the field varies only along z.
So,
F = p × (dE/dz)

Now substitute values:

F = 10⁻⁷ C·m × 10⁵ N/C/m
F = 0.01 N
The direction is along the negative z-axis (because the dipole moment is in that direction).

Torque on the Dipole:

Torque (τ) is given by:
τ = p × E

Since both p and E are along the z-direction, the angle between them is either 0° or 180°, so:

τ = 0

No torque acts on the dipole in this case.

Final Answer:

• Force on the dipole = 0.01 N in the negative z-direction
• Torque = 0

Question 1.28:
(a) A conductor A with a cavity as shown in Figure (a) is given a charge
Q. Show that the entire charge must appear on the outer surface of the
conductor.
(b) Another conductor B with charge q is inserted into the cavity keeping
B insulated from A. Show that the total charge on the outside surface of A is
Q + q [Figure (b)].
(c) A sensitive instrument is to be shielded from the strong electrostatic
fields in its environment. Suggest a possible way.

(a) Entire charge Q appears on the outer surface of conductor A:

• A conductor in electrostatic equilibrium has zero electric field inside its bulk.
• If any charge remained on the inner surface or inside the conductor material, it would create an internal electric field, violating electrostatic equilibrium.
• Hence, all charge Q must reside on the outer surface of conductor A.

(b) Conductor B with charge q placed inside the cavity:

• When conductor B with charge q is inserted into the cavity (without touching A), it induces an equal and opposite charge (−q) on the inner surface of A (facing the cavity) to maintain zero field inside the conductor material of A.
• Since conductor A already had a total charge Q, and now −q appears on the inner surface, the remaining charge on the outer surface of A must be: Outer charge = Q − (−q) = Q + q

Hence, total charge on the outside surface of A becomes Q + q.

(c) Shielding a sensitive instrument:

• To protect a sensitive instrument from external electric fields, place it inside a hollow conducting shell (like a metal box).
• This method is known as electrostatic shielding.
• The conductor redistributes its charges in such a way that electric field inside the cavity is zero, regardless of external fields.

This principle is used in Faraday cages, which block external electric fields completely.

Final Answer Summary:

• (a) All charge Q resides on the outer surface of conductor A.
• (b) When a charge q is inserted inside the cavity, outer surface charge becomes Q + q.
• (c) Use a hollow conductor (Faraday cage) to shield sensitive instruments from electric fields.

Question 1.30:
Obtain the formula for the electric field due to a long thin wire of uniform
linear charge density λ without using Gauss’s law. [Hint: Use Coulomb’s law
directly and evaluate the necessary integral.]
Answer 1.30:
Let’s consider a long thin wire placed along the x-axis with a uniform linear charge density λ (lambda), which means charge per unit length.

We want to find the electric field at a point located a perpendicular distance r from the wire along the y-axis (at the point (0, r)).

Step 1: Choose a small element of the wire

Take a small length element dx at a position x on the x-axis.
The charge of this small element is:
dq = λ dx

Step 2: Use Coulomb’s Law

According to Coulomb’s Law, the electric field dE at point P due to this element is:

dE = (1 / (4πϵ₀)) * (dq / (x² + r²)) * (unit vector)

The distance from the element to point P is:
sqrt(x² + r²)

The direction (unit vector) from the element to P is:
(-x î + r ĵ) / sqrt(x² + r²)

So,
dE = (λ / (4πϵ₀)) * (-x î + r ĵ) dx / (x² + r²)^(3/2)

Step 3: Integrate over the wire

To find the total electric field, integrate from x = -∞ to x = ∞.

X-component (Ex):
Ex = (λ / (4πϵ₀)) * ∫[-∞ to ∞] (-x / (x² + r²)^(3/2)) dx
This integral is an odd function over symmetric limits, so:
Ex = 0

Y-component (Ey):
Ey = (λ / (4πϵ₀)) * ∫[-∞ to ∞] (r / (x² + r²)^(3/2)) dx
Take r outside:
Ey = (λ * r / (4πϵ₀)) * ∫[-∞ to ∞] (1 / (x² + r²)^(3/2)) dx

This standard integral evaluates to:
∫[-∞ to ∞] (1 / (x² + r²)^(3/2)) dx = 2 / r²

So:
Ey = (λ * r / (4πϵ₀)) * (2 / r²) = λ / (2πϵ₀r)

Final Result:

The total electric field at a distance r from the wire is:

E = λ / (2πϵ₀r)

This field is directed radially outward from the wire (if λ is positive), and is perpendicular to the wire.

Question 1.31:
It is now believed that protons and neutrons (which constitute nuclei of
ordinary matter) are themselves built out of more elementary units called
quarks. A proton and a neutron consist of three quarks each. Two types of
quarks, the so called ‘up’ quark (denoted by u) of charge (+2/3) e, and the
‘down’ quark (denoted by d) of charge (−1/3) e, together with electrons build
up ordinary matter. (Quarks of other types have also been found which give
rise to different unusual varieties of matter.) Suggest a possible quark
composition of a proton and neutron.

Answer 1.31

Quark Composition of Proton and Neutron

Protons and neutrons, which make up the nuclei of atoms, are not fundamental particles. They are composed of even more basic particles known as quarks.

There are different types of quarks, but only two types are needed to explain the structure of ordinary matter:

• Up quark (u): has a charge of (+2/3) e
• Down quark (d): has a charge of (−1/3) e

Each proton or neutron is made up of three quarks.

Quark Composition of a Proton

A proton has a total charge of +1 e.

Let’s consider a combination of quarks that gives this charge:

• 2 up quarks → 2 × (+2/3) e = +4/3 e
• 1 down quark → (−1/3) e

Total charge = (+4/3) + (−1/3) = +1 e

So, the quark composition of a proton is:

Proton = u + u + d

Quark Composition of a Neutron

A neutron has a total charge of 0 e.

Let’s consider a combination of quarks that gives zero total charge:

• 1 up quark → (+2/3) e
• 2 down quarks → 2 × (−1/3) e = −2/3 e

Total charge = (+2/3) + (−2/3) = 0 e

So, the quark composition of a neutron is:

Neutron = u + d + d

Summary

• Proton = up + up + down (u u d) → charge = +1 e
• Neutron = up + down + down (u d d) → charge = 0 e

These combinations of quarks form the building blocks of all ordinary atomic nuclei.

Question 1.32:
(a) Consider an arbitrary electrostatic field configuration. A small test
charge is placed at a null point (i.e., where E = 0) of the configuration. Show
that the equilibrium of the test charge is necessarily unstable.
(b) Verify this result for the simple configuration of two charges of the
same magnitude and sign placed a certain distance apart.
Answer 1.32:
(a) Let the equilibrium of the test charge be stable. If a test charge is in
equilibrium and displaced from its position in any direction, then it
experiences a restoring force towards a null point, where the electric field is
zero.

All the field lines near the null point are directed inwards towards the
null point. There is a net inward flux of electric field through a closed surface
around the null point. According to Gauss’s law, the flux of electric field
through a surface, which is not enclosing any charge, is zero. Hence, the
equilibrium of the test charge can be stable.

(b) Two charges of same magnitude and same sign are placed at a certain
distance. The mid-point of the joining line of the charges is the null point.
When a test charged is displaced along the line, it experiences a restoring
force. If it is displaced normal to the joining line, then the net force takes it
away from the null point. Hence, the charge is unstable because stability of
equilibrium requires restoring force in all directions.

Question 1.33
A particle of mass m and charge (−q) enters the region between the two
charged plates initially moving along x-axis with speed vx (like particle 1 in
Figure). The length of plate is L and an uniform electric field E is maintained
between the plates. Show that the vertical deflection of the particle at the far
edge of the plate is qEL2/(2m𝑣𝑥2).
Compare this motion with motion of a projectile in gravitational field
Answer 1.33

Vertical Deflection of a Charged Particle in an Electric Field Between Plates

Let a particle of mass m and charge −q enter the region between two charged plates.

• It enters horizontally along the x-axis with an initial speed vₓ.
• The length of the plates is L.
• A uniform electric field E is present vertically upward between the plates.

We are asked to find the vertical deflection y of the particle as it exits the plates (i.e., when it has traveled a horizontal distance L).

Step-by-Step Derivation

1. Horizontal Motion (along x-axis):

• There is no force acting in the horizontal direction.
• So, the horizontal velocity remains constant at vₓ.
• Time taken to cross the plates of length L is: t = L / vₓ

2. Vertical Motion (along y-axis):

• The particle has charge −q.
• The electric field E is upward → force on the particle is downward (because the charge is negative): F = −qE (direction: downward)
• The vertical acceleration a is: a = F / m = −qE / m
• Initial vertical velocity is 0 (since particle enters horizontally).
• Using the equation of motion: y = 0 * t + (1/2) * a * t²
y = (1/2) * (−qE / m) * (L / vₓ)² y = −(qE * L²) / (2m * vₓ²)

So, the vertical deflection at the far edge is:

iniCopyEdity = −qEL² / (2mvₓ²)

The negative sign indicates that the deflection is downward.

Comparison with Projectile Motion in a Gravitational Field

This motion is very similar to the motion of a projectile:

• In projectile motion:
  • Horizontal motion: constant velocity
  • Vertical motion: uniform acceleration due to gravity g
• In this electric field case:
  • Horizontal motion: constant velocity vₓ
  • Vertical motion: uniform acceleration a = −qE/m (acts like “electric gravity”)

So, the electric field plays the same role in this case as gravity does in projectile motion.

Just like in projectile motion:

• Time of flight depends on horizontal velocity
• Vertical displacement depends on acceleration and time squared

Conclusion

The motion of a charged particle in a uniform electric field between plates is analogous to a projectile motion under gravity.

• Vertical deflection of the particle is: y = −qEL² / (2mvₓ²)
• The trajectory is parabolic, just like in projectile motion.

Question 1.34
Suppose that the particle in Exercise in 1.33 is an electron projected with
velocity vx= 2.0 × 106 m s−1. If E between the plates separated by 0.5 cm is
9.1 × 102 N/C, where will the electron strike the upper plate?
(| e | =1.6 × 10−19 C, me = 9.1 × 10−31 kg.)
Answer 1.34:

Given:

• Initial horizontal velocity,
  vₓ = 2.0 × 10⁶ m/s
• Electric field,
  E = 9.1 × 10² N/C
• Separation between plates,
  d = 0.5 cm = 0.005 m
• Charge of electron,
  q = 1.6 × 10⁻¹⁹ C (electron has negative charge, so direction matters)
• Mass of electron,
  m = 9.1 × 10⁻³¹ kg

Step-by-Step Calculation:

1. Vertical acceleration due to electric field

For an electron in an electric field,
a = F / m = (−eE) / m

So,

a = − (1.6 × 10⁻¹⁹ × 9.1 × 10²) / (9.1 × 10⁻³¹)

Let’s compute this:

a = − (1.6 × 9.1 × 10⁻¹⁷) / (9.1 × 10⁻³¹)
a = − (1.6 × 10⁻¹⁷) / (10⁻³¹)
a = − 1.6 × 10¹⁴ m/s²

So, vertical acceleration:

a = −1.6 × 10¹⁴ m/s²

2. Time taken to cross the plates (horizontal direction)

We don’t know the length L, but we are told to find where the electron strikes the upper plate. That means we need to find when vertical displacement equals the plate separation.

So,

Let vertical displacement y = +0.005 m (upward, since field is upward and electron is pulled downward due to negative charge).

We use:

y = (1/2) a t²

Substitute values:

0.005 = (1/2) * 1.6 × 10¹⁴ * t²

Solving:

t² = (2 × 0.005) / (1.6 × 10¹⁴)
t² = 0.01 / (1.6 × 10¹⁴)
t² = 6.25 × 10⁻¹٧
t = √(6.25 × 10⁻¹٧) = 2.5 × 10⁻⁹ s

3. Horizontal distance traveled before striking the plate

Use:

x = vₓ * t = (2.0 × 10⁶) * (2.5 × 10⁻⁹)
x = 5.0 × 10⁻³ m = 0.005 m = 0.5 cm

Final Answer:

The electron will strike the upper plate after traveling 0.5 cm horizontally, which means it just hits the upper plate exactly at the exit point of the plates if their length is also 0.5 cm.

Conclusion: Class 12 Physics Ch Electric Charges and Fields

In this chapter Class 12 Physics Ch Electric Charges and Fields, we explored the fundamental concepts of electric charges and electric fields, which are essential in understanding the behavior of charged particles in different environments. The study of electric forces, Coulomb’s law, and the electric field concept provides the foundation for analyzing the interactions between charged objects. By studying electric fields, we are able to comprehend the behavior of charged particles and the effects of external fields on them.

In conclusion, the study of electric fields and forces not only enhances our theoretical understanding but also prepares us to explore more complex concepts in physics. From everyday applications to advanced scientific research, the electric field plays a pivotal role in shaping our understanding of the universe.

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