Question 1

Q1. The dielectric constant of an insulator cannot be:

Accepted Answer

∝

Question 2

Q2. A body can be negatively charged

Accepted Answer

By adding some electrons to it

Question 3

Q3.The electric field at a distance R due to charge q is E. If the same charge is placed on the copper spherical shell of radius R, the electric field strength at the surface of the conductor will be

Accepted Answer

E

Question 4

Q4.In a region where intensity of electric field is 5NC -1 , 40 electric field lines are crossing per square metre. The number of electric field lines crossing per square metre where intensity of electric field is 10 NC -1 , will be

Accepted Answer

80

Question 5

Q5 If an electric dipole is placed in a uniform electric field , it experiences

Accepted Answer

Torque only

Question 6

Two point charges +10 µC and -10 µC are separated by a distance of 2 cm in water of dielectric constant 80. The potential energy of the system is-

Accepted Answer

-0.56 J

Question 7

Which of the following angle has minimum potential energy when dipole is placed in uniform electric field

Accepted Answer

0º

Question 8

What is the force between two small charged spheres having charges of 2 × 10–7C and 3 × 10–7C placed 30 cm apart in air?

Accepted Answer

Given: q1 = 2 × 10⁻⁷ C, q2 = 3 × 10⁻⁷ C, r = 30 cm = 0.3 m

Using Coulomb's law:
F = k * |q1 * q2| / r²
where k = 9 × 10⁹ N m²/C²

F = 9 × 10⁹ × (2 × 10⁻⁷) × (3 × 10⁻⁷) / (0.3)²
= 9 × 10⁹ × 6 × 10⁻¹⁴ / 0.09
= (9 × 6 × 10⁻⁵) / 0.09
= (54 × 10⁻⁵) / 0.09
= 6 × 10⁻³ N

Therefore, the force between the spheres is 6 × 10⁻³ N. — Step 1: Identify given charges and distance.
Step 2: Use Coulomb's law formula.
Step 3: Substitute values and calculate.
Step 4: Simplify to get the force magnitude.

Question 9

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?

Accepted Answer

(a) Given: q1 = 0.4 mC = 0.4 × 10⁻³ C, q2 = –0.8 mC = –0.8 × 10⁻³ C, F = 0.2 N

Using Coulomb's law:
F = k * |q1 * q2| / r²
=> r² = k * |q1 * q2| / F

Substitute values:
r² = (9 × 10⁹) × (0.4 × 10⁻³) × (0.8 × 10⁻³) / 0.2
= 9 × 10⁹ × 0.32 × 10⁻⁶ / 0.2
= (2.88 × 10³) / 0.2 = 1.44 × 10⁴

r = √(1.44 × 10⁴) = 120 m

(b) By Newton's third law, the force on the second sphere due to the first is equal in magnitude and opposite in direction to the force on the first sphere due to the second.

Therefore, force on second sphere = 0.2 N (magnitude), direction opposite to force on first sphere. — Step 1: Use given force and charges to find distance using Coulomb's law.
Step 2: Rearrange formula to solve for r.
Step 3: Apply Newton's third law for force on second sphere.

Question 10

Check that the ratio ke2/G m m is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

Accepted Answer

Let us consider the ratio $ \frac{k e^2}{G m_e m_p} $, where:
- k = Coulomb's constant = 9 × 10⁹ N m²/C²
- e = elementary charge = 1.6 × 10⁻¹⁹ C
- G = gravitational constant = 6.67 × 10⁻¹¹ N m²/kg²
- m_e = mass of electron = 9.11 × 10⁻³¹ kg
- m_p = mass of proton = 1.67 × 10⁻²⁷ kg

First, check dimensions:
- k has units N m²/C² = (kg m/s²) m² / C² = kg m³ / (s² C²)
- e² has units C²
- So, k e² has units kg m³ / s²
- G has units N m²/kg² = (kg m/s²) m² / kg² = m³ / (s² kg)
- m_e m_p has units kg²
- So, G m_e m_p has units m³ / s²

Therefore, $ \frac{k e^2}{G m_e m_p} $ is dimensionless.

Calculate value:

Numerator: k e² = 9 × 10⁹ × (1.6 × 10⁻¹⁹)² = 9 × 10⁹ × 2.56 × 10⁻³⁸ = 2.304 × 10⁻²⁸

Denominator: G m_e m_p = 6.67 × 10⁻¹¹ × 9.11 × 10⁻³¹ × 1.67 × 10⁻²⁷ = 6.67 × 10⁻¹¹ × 1.52 × 10⁻⁵7 = 1.01 × 10⁻⁶7

Ratio = 2.304 × 10⁻²⁸ / 1.01 × 10⁻⁶7 ≈ 2.3 × 10³9

This large dimensionless number signifies that the electrostatic force between an electron and a proton is about 10³9 times stronger than the gravitational force between them. — Step 1: Analyze units of numerator and denominator to confirm dimensionless.
Step 2: Substitute known physical constants.
Step 3: Calculate numerical value.
Step 4: Interpret the physical significance of the ratio.

Question 11

(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?

Accepted Answer

(a) The statement 'electric charge of a body is quantised' means that the charge on any body is always an integral multiple of the elementary charge e (charge of an electron or proton). That is, charge q = n e, where n is an integer.

(b) Quantisation can be ignored in macroscopic charges because the number of elementary charges involved is extremely large, making the charge appear continuous. The smallest unit e is so small that for large charges, the discrete nature is negligible. — Step 1: Define quantisation of charge.
Step 2: Explain why in large scale charges, the discrete nature is not noticeable.

Question 12

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.

Accepted Answer

When a glass rod is rubbed with a silk cloth, electrons are transferred from one to the other. For example, electrons move from the glass rod to the silk cloth, making the rod positively charged and the cloth negatively charged. The total charge before and after rubbing remains the same (zero net charge), only redistributed. This illustrates the law of conservation of charge which states that the total charge in an isolated system remains constant. — Step 1: Describe charge transfer during rubbing.
Step 2: Show that total charge is conserved by redistribution.
Step 3: Conclude consistency with conservation law.

Question 13

Four point charges q = 2 mC, q = –5 mC, q = 2 mC, and q = –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?

Accepted Answer

Given:
Charges at corners: q_A = 2 mC, q_B = –5 mC, q_C = 2 mC, q_D = –5 mC
Side of square, a = 10 cm = 0.1 m
Charge at center, q_0 = 1 mC = 1 × 10⁻³ C

Distance from center to each corner = (a/√2) = 0.1 / 1.414 = 0.0707 m

Calculate force due to each corner charge on q_0 using Coulomb's law:
F = k |q q_0| / r²

Calculate magnitude of force from each charge:
F_A = 9 × 10⁹ × 2 × 10⁻³ × 1 × 10⁻³ / (0.0707)² = 9 × 10⁹ × 2 × 10⁻⁶ / 0.005 = 3.6 × 10⁶ N

Similarly, F_B = 9 × 10⁹ × 5 × 10⁻³ × 1 × 10⁻³ / 0.005 = 9 × 10⁹ × 5 × 10⁻⁶ / 0.005 = 9 × 10⁶ N

Directions:
Charges 2 mC are positive, so force on positive q_0 is repulsive.
Charges –5 mC are negative, so force on positive q_0 is attractive.

Due to symmetry, forces from opposite corners add vectorially.

Calculate net force vector by adding components.

Result:
The net force on the 1 mC charge at the center is zero because the forces due to charges at opposite corners cancel out due to symmetry and equal magnitude but opposite sign charges.

Therefore, net force = 0 N. — Step 1: Calculate distance from center to corners.
Step 2: Calculate magnitude of force from each corner charge.
Step 3: Determine direction of forces.
Step 4: Use symmetry to add forces vectorially.
Step 5: Conclude net force is zero.

Question 14

(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?

Accepted Answer

(a) Electric field lines represent the direction of the electric field vector at every point. Since the electric field is continuous in space (except at point charges), the field lines must also be continuous curves without breaks.

(b) Two field lines never cross because at the point of crossing, the electric field would have two different directions simultaneously, which is impossible. The electric field vector at any point has a unique direction. — Step 1: Explain continuity of electric field and hence field lines.
Step 2: Explain uniqueness of electric field direction preventing crossing.

Question 15

Two point charges q = 3 mC and q = –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?

Accepted Answer

(a) Given:
q_A = 3 mC = 3 × 10⁻³ C
q_B = –3 mC = –3 × 10⁻³ C
Distance AB = 20 cm = 0.2 m
Midpoint O is 0.1 m from each charge.

Electric field due to q_A at O:
E_A = k |q_A| / r² = 9 × 10⁹ × 3 × 10⁻³ / (0.1)² = 9 × 10⁹ × 3 × 10⁻³ / 0.01 = 2.7 × 10⁹ N/C
Direction: away from positive charge q_A towards O.

Electric field due to q_B at O:
E_B = k |q_B| / r² = same magnitude = 2.7 × 10⁹ N/C
Direction: towards negative charge q_B, i.e., also towards O.

Since both fields at O point in the same direction (from q_A to q_B), total field:
E_total = E_A + E_B = 2 × 2.7 × 10⁹ = 5.4 × 10⁹ N/C

(b) Force on test charge q_t = –1.5 × 10⁻⁹ C:
F = q_t × E_total = (–1.5 × 10⁻⁹) × 5.4 × 10⁹ = –8.1 N

The negative sign indicates force direction opposite to field direction, i.e., towards q_A.

Therefore, magnitude of force = 8.1 N directed towards q_A. — Step 1: Calculate electric field at midpoint due to each charge.
Step 2: Add vectorially since fields are in same direction.
Step 3: Calculate force on test charge using F = qE.
Step 4: Interpret sign and direction of force.

Electric Charges and Fields

Electric Charges and Fields — Study Notes

Introduction

Electric Charge

Coulomb's Law

Practice Questions — Electric Charges and Fields

All 8 Chapters in Physics Part-I