ELECTRIC CAPACITY - XII Notes- with numericals

 ELECTRIC   CAPACITY

IF the amount of water is added in any pot then the level of liquid in the container increases and water will be accumulated in the container up to  a certain level after wards it will  overflow.               similarly if change  is given to any conductor then it’ s  level of charging or say potential will increase . More the charge more the potential.

         Thus we can conclude potential of any conductor and amount of change present on the conductor and proportional to each other. If we give  ‘Q’ change to any conductor and increase in it’s potential is V then  

                        Q     V          

                        Q =VC              Where,  C is the electric capacity of the conductor .  or Capacitance of the

  conductor of the conductor

    

        CAPACITY OF ANY CONDUCTOR   IS CONSTANT WHICH DEPENDS ON -->

          (1)  Shape of the conductor                  (3) Medium around the conductor.

(2)  Area of the conductor.                   (4) Presence of any other conductor close to the charged conductor

                                                              Q=CV

                         (i)                                  C=Q/V                          

                         (ii)     If   V =Unit increase in potential

                                                           

                                                             C=Q             

 DEFINITION OF ELECTRIC CAPACITY =>         

                                        Electric capacity of any conductor is the ratio of charge given to the conductor and increase in the potential of the conductor. OR .    Electric capacity of any conductor is numerically equal to the amount of charge required to increase its  potential by unit.

   Unit of capacity—       C=Q/V                                     1 Farad =10⁶  Micro Farad      1 Micro Farad= 10^-6 Farad                  

                                                                                        1 Farad=10¹²  Pico Farad        1 Pico Farad =10^-12 Farad                             

                                       

Definition of One   Farad =>

If the increase in potential of any conductor is one volt due to the charge of one  coulomb then the capacity of the conductor will be one Farad .

CAPACITY OF ANY ISOLATED SPERICAL CONDUCTOR 

    Consider any spherical conductor having radius ‘R’ center ‘O’ and charge given will be uniformly distributed over the surface of the conductor so its surface will be equipotent surface and electric lines of force will be along the radius now potential on the conductor will be  V-

Atmospheric electricity - class 12 notes with numercals

Atmospheric Electricity: Phenomena, Measurements, and Applications.

Atmospheric Electricity: Phenomena, Measurements, and Applications.

Atmospheric electricity refers to the electrical properties and phenomena that occur in the Earth's atmosphere. It is a complex and fascinating field of study that has captured the attention of scientists for over a century. The Earth's atmosphere is constantly charged due to the interaction of the Sun's radiation, cosmic rays, and thunderstorm activity. This has led to the discovery of many important phenomena, including lightning, the ionosphere, and the aurora borealis. Atmospheric electricity has practical applications in areas such as weather forecasting, radio communication, and the design of lightning protection systems. In this article, we will explore the sources of atmospheric electricity, the different atmospheric electrical phenomena, the measurements used to study it, and its practical applications. We will also answer some frequently asked questions about atmospheric electricity

Table of Contents: Atmospheric Electricity: Phenomena, Measurements, and Applications.

• Introduction to Atmospheric Electricity

• Sources of Atmospheric Electricity

• Atmospheric Electrical Phenomena

• Lightning

• Thunderstorms

• Atmospheric Electrical Measurements

• Applications of Atmospheric Electricity

• Frequently Asked Questions

1. Introduction to Atmospheric Electricity

Atmospheric electricity refers to the electrical properties and phenomena that occur in the Earth's atmosphere. The Earth's atmosphere is constantly charged due to the interaction of the Sun's radiation, cosmic rays, and thunderstorm activity. Atmospheric electricity has been studied for over a century and has led to the discovery of many important phenomena, including lightning, the ionosphere, and the aurora borealis.

2. Sources of Atmospheric Electricity

The primary sources of atmospheric electricity are the Sun's radiation, cosmic rays, and thunderstorm activity. The Sun's radiation causes the ionization of gases in the upper atmosphere, leading to the formation of the ionosphere. Cosmic rays, which are high-energy particles from space, also contribute to the ionization of the atmosphere. Thunderstorm activity causes the buildup of electric charges in the lower atmosphere, leading to lightning and other electrical phenomena.

3. Atmospheric Electrical Phenomena

Atmospheric electrical phenomena include lightning, thunderstorms, sprites, and elves. Lightning is the most well-known and dramatic atmospheric electrical phenomenon. It is caused by the buildup of electric charges in the atmosphere, leading to a discharge of electricity between the clouds and the ground. Thunderstorms are also a common atmospheric electrical phenomenon and are characterized by the buildup of electric charges in the lower atmosphere, leading to thunder and lightning.

Sprites and elves are less well-known atmospheric electrical phenomena. Sprites are large-scale electrical discharges that occur high in the atmosphere, above thunderstorms. Elves are electromagnetic pulses that occur in the ionosphere, caused by the interaction of lightning discharges with the ionosphere.

4. Lightning

Lightning is a natural electrical discharge that occurs in the atmosphere. It is caused by the buildup of electric charges in the atmosphere, which leads to a discharge of electricity between the clouds and the ground. Lightning can be very dangerous and can cause injury or death to people and damage to buildings and other structures.

There are several types of lightning, including cloud-to-ground lightning, intra-cloud lightning, and cloud-to-cloud lightning. Cloud-to-ground lightning is the most well-known type and occurs when a discharge of electricity travels from the cloud to the ground. Intra-cloud lightning occurs entirely within the cloud, while cloud-to-cloud lightning occurs between different clouds.

5. Thunderstorms

Thunderstorms are a type of weather phenomenon characterized by the buildup of electric charges in the atmosphere. Thunderstorms can produce lightning, thunder, heavy rain, strong winds, and even tornadoes. Thunderstorms can be very dangerous and can cause property damage and loss of life.

Thunderstorms are caused by the interaction of warm and cold air masses. Warm air rises, and as it does, it cools and condenses into clouds. The updrafts within the cloud cause the buildup of electric charges, which can lead to lightning and thunder.

6. Atmospheric Electrical Measurements

Atmospheric electrical measurements are used to study the electrical properties of the atmosphere. These measurements include electric field measurements, atmospheric conductivity measurements, and measurements of atmospheric ionization.

Electric field measurements are used to measure the strength and direction of the electric field in the atmosphere. Atmospheric conductivity measurements are used to measure the ability of the atmosphere to conduct electricity. Measurements of atmospheric ionization are used to measure the concentration of ions in the atmosphere.

7. Applications of Atmospheric Electricity

Atmospheric electricity has many practical applications. For example, it is used in the design of lightning protection systems for buildings and other structures. Atmospheric electricity is also used in the study of the ionosphere, which is important for radio communication and satellite navigation. Additionally, atmospheric electrical measurements can be used to study weather patterns and predict severe weather events.

The earth’s atmosphere extends to about 300 Km above the earth surface. The atmosphere is divided in four layers.

300 km               400°C                                                                                     d4 = d/1010 Ionosphere                               Good Conductor

80 km               -90°C                                                                                     d3 = d/105 Mesosphere

50 Km               10°C                                                                                      d2 = d/1000 Stratosphere

12 km                -50°C                                                                                    d1 = d/10 Troposphere                             Poor Conductor                       Temp = 15°C                                              density of air d = 1.29kg/m3

Electrical properties of the atmosphere: -

(1)    The electrical phenomena in atmosphere take place between the earth surface and top of stratosphere. The 50 Km thick layers is like a blanket enveloping the earth.

(2)  An electric field 100 V/m is there downwards all over the earth, at ground level.

8. Frequently Asked Questions: FAQs

Q: Can atmospheric electricity be used as a source of energy? A: While atmospheric electricity is a source of energy, it is not currently practical to harness it for energy production due to the high cost and low efficiency of existing technologies.

Q: How is lightning formed? A: Lightning is formed by the buildup of electric charges in the atmosphere, which leads to a discharge of electricity between the clouds and the ground.

Q: What is the ionosphere? A: The ionosphere is a region of the Earth's upper atmosphere that is ionized by solar radiation. It plays an important role in radio communication and satellite navigation.

Q: What is atmospheric conductivity? A: Atmospheric conductivity is a measure of the ability of the atmosphere to conduct electricity. It is influenced by factors such as temperature, humidity, and ionization.

Q: How can atmospheric electricity be measured? A: Atmospheric electricity can be measured using a variety of instruments, including electric field meters, ion counters, and atmospheric conductivity meters. These instruments are used to measure the electric field, ionization, and conductivity of the atmosphere.

Q: Is atmospheric electricity dangerous? A: Atmospheric electricity can be dangerous, particularly during thunderstorms and lightning strikes. It is important to take precautions to protect yourself during severe weather events.

Q: What are sprites and elves? A: Sprites and elves are atmospheric electrical phenomena that occur high in the atmosphere. Sprites are large-scale electrical discharges, while elves are electromagnetic pulses caused by the interaction of lightning with the ionosphere.

          

GRAVITATION : Introduction, Detailed description, numericals, FAQs

Gravitations introduction, detailed explanation with equations, numericals, FAQs

Introduction:

Gravity is the force by which a planet or other body draws objects toward its center. The force of gravity keeps all of the planets in orbit around the sun. Earth's gravity is what keeps you on the ground and what makes things fall.

Detailed Explanation with Equations:

Gravity is described by the law of universal gravitation, which states that every point mass attracts every other point mass by a force acting along the line intersecting both points. The force is proportional to the product of the two masses and inversely proportional to the square of the distance between them. The equation for the law of universal gravitation is given by:

F = G (m1m2/r²)

Where F is the force of gravity between the two objects, m1 and m2 are the masses of the two objects, r is the distance between the two objects, and G is the gravitational constant.

The gravitational constant, denoted by G, is a fundamental constant of nature that appears in the law of universal gravitation. Its value is approximately 6.674 x 10^-11 N·m²/kg².

The force of gravity is always attractive, which means that it pulls objects together. The magnitude of the force depends on the masses of the objects and the distance between them. If the masses of the objects are large, the force of gravity between them is also large. If the distance between the objects is large, the force of gravity between them is weaker.

Numericals:

Here are a few numerical examples to illustrate the use of the law of universal gravitation:

• Calculate the force of gravity between two objects with masses of 10 kg and 20 kg that are separated by a distance of 5 meters.

Solution:

F = G (m1m2/r²) F = 6.674 x 10^-11 N·m²/kg² (10 kg x 20 kg)/(5 m)² F = 2.004 x 10^-9 N

Therefore, the force of gravity between the two objects is 2.004 x 10^-9 N.

• A 1,000-kg satellite is in orbit around Earth at an altitude of 500 km. What is the force of gravity on the satellite?

Solution:

The distance between the satellite and the center of Earth is equal to the sum of the radius of Earth (6,371 km) and the altitude of the satellite (500 km).

r = 6,371 km + 500 km = 6,871 km = 6,871,000 meters

The mass of Earth is 5.97 x 10^24 kg.

F = G (m1m2/r²) F = 6.674 x 10^-11 N·m²/kg² (1,000 kg x 5.97 x 10^24 kg)/(6,871,000 m)² F = 8,869 N

Therefore, the force of gravity on the satellite is 8,869 N.

Problem: A planet of mass 6 x 10^24 kg has a radius of 6.4 x 10^6 m. A satellite of mass 1000 kg is orbiting around it at a height of 8000 km above the surface of the planet. Calculate: (a) The acceleration due to gravity on the surface of the planet. (b) The speed of the satellite in its orbit. (c) The time period of the satellite's orbit.

Solution: (a) The acceleration due to gravity on the surface of the planet can be calculated using the formula:

g = G M / R^2

where g is the acceleration due to gravity, G is the universal gravitational constant, M is the mass of the planet, and R is the radius of the planet.

Plugging in the given values, we get:

g = (6.67 x 10^-11 Nm^2/kg^2) x (6 x 10^24 kg) / (6.4 x 10^6 m)^2 g = 9.81 m/s^2

Therefore, the acceleration due to gravity on the surface of the planet is 9.81 m/s^2.

(b) The speed of the satellite in its orbit can be calculated using the formula:

v = √(G M / r)

where v is the speed of the satellite, G is the universal gravitational constant, M is the mass of the planet, and r is the distance of the satellite from the center of the planet.

The distance of the satellite from the center of the planet is:

r = R + h = (6.4 x 10^6 m) + (8000 km) = 6.4 x 10^6 m + 8 x 10^6 m = 1.44 x 10^7 m

Plugging in the given values, we get:

v = √[(6.67 x 10^-11 Nm^2/kg^2) x (6 x 10^24 kg) / (1.44 x 10^7 m)] v = 10,968 m/s (approx.)

Therefore, the speed of the satellite in its orbit is approximately 10,968 m/s.

(c) The time period of the satellite's orbit can be calculated using the formula:

T = 2πr / v

where T is the time period of the orbit, r is the distance of the satellite from the center of the planet, and v is the speed of the satellite.

Plugging in the values we have calculated, we get:

T = 2π x (1.44 x 10^7 m) / (10,968 m/s) T = 13,500 seconds (approx.)

Therefore, the time period of the satellite's orbit is approximately 13,500 seconds or 3.75 hours.

Problem: Two objects of masses 10 kg and 5 kg are placed at a distance of 2 m from each other. Calculate the gravitational force of attraction between them. Also, calculate the acceleration due to gravity experienced by each of the objects due to this force.

Solution: The gravitational force of attraction between the two objects can be calculated using the formula:

F = G (m1m2) / r^2

where F is the force of attraction, G is the universal gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between the centers of the two objects.

Plugging in the given values, we get:

F = (6.67 x 10^-11 Nm^2/kg^2) x (10 kg) x (5 kg) / (2 m)^2 F = 8.34 x 10^-10 N

Therefore, the gravitational force of attraction between the two objects is 8.34 x 10^-10 N.

The acceleration due to gravity experienced by each of the objects due to this force can be calculated using the formula:

a = F / m

where a is the acceleration due to gravity, F is the gravitational force of attraction, and m is the mass of the object.

For the 10 kg mass, the acceleration due to gravity is:

a = (8.34 x 10^-10 N) / (10 kg) a = 8.34 x 10^-11 m/s^2

For the 5 kg mass, the acceleration due to gravity is:

a = (8.34 x 10^-10 N) / (5 kg) a = 1.67 x 10^-9 m/s^2

Therefore, the acceleration due to gravity experienced by the 10 kg object is 8.34 x 10^-11 m/s^2 and the acceleration due to gravity experienced by the 5 kg object is 1.67 x 10^-9 m/s^2.

• Two objects of masses 5 kg and 10 kg are placed at a distance of 4 m from each other. Calculate the gravitational force of attraction between them.

Solution: Using the formula F = G (m1m2) / r^2, we get:

F = (6.67 x 10^-11 Nm^2/kg^2) x (5 kg) x (10 kg) / (4 m)^2 F = 8.34 x 10^-11 N

Therefore, the gravitational force of attraction between the two objects is 8.34 x 10^-11 N.

• A planet has a mass of 6 x 10^24 kg and a radius of 6.4 x 10^6 m. Calculate the acceleration due to gravity on the surface of the planet.

Solution: Using the formula g = G M / R^2, we get:

g = (6.67 x 10^-11 Nm^2/kg^2) x (6 x 10^24 kg) / (6.4 x 10^6 m)^2 g = 9.81 m/s^2

Therefore, the acceleration due to gravity on the surface of the planet is 9.81 m/s^2.

• A satellite of mass 1000 kg is in orbit around a planet with a mass of 6 x 10^24 kg and a radius of 6.4 x 10^6 m. The satellite is orbiting at a height of 8000 km above the surface of the planet. Calculate the speed of the satellite in its orbit.

Solution: The distance of the satellite from the center of the planet is:

r = R + h = (6.4 x 10^6 m) + (8000 km) = 6.4 x 10^6 m + 8 x 10^6 m = 1.44 x 10^7 m

Using the formula v = √(G M / r), we get:

v = √[(6.67 x 10^-11 Nm^2/kg^2) x (6 x 10^24 kg) / (1.44 x 10^7 m)] v = 10,968 m/s (approx.)

Therefore, the speed of the satellite in its orbit is approximately 10,968 m/s.

• A planet has a mass of 2 x 10^23 kg and a radius of 4000 km. Calculate the escape velocity from the surface of the planet.

Solution: Using the formula v = √(2GM/R), we get:

v = √[(2 x 6.67 x 10^-11 Nm^2/kg^2 x 2 x 10^23 kg) / (4000 km + 6.4 x 10^6 m)] v = 4.18 km/s (approx.)

Therefore, the escape velocity from the surface of the planet is approximately 4.18 km/s.

• Two objects of masses 2 kg and 4 kg are placed at a distance of 3 m from each other. Calculate the gravitational potential energy of the system.

Solution: Using the formula U = -G(m1m2)/r, we get:

U = -(6.67 x 10^-11 Nm^2/kg^2) x (2 kg) x (4 kg) / (3 m) U = -1.78 x 10^-10 J

Therefore, the gravitational potential energy of the system is -1.78 x 10^-10 J.

• A planet of mass 3 x 10^24 kg has a moon of mass 5 x 10^22 kg orbiting around it in a circular orbit at a distance of 4 x 10^5 km. Calculate the speed of the moon in its orbit.

Solution: Using the formula v = √(GM/r), we get:

v = √[(6.67 x 10^-11 Nm^2/kg^2 x 3 x 10^24 kg) / (4 x 10^5 km + 6.4 x 10^6 m)] v = 1024 m/s (approx.)

Therefore, the speed of the moon in its orbit is approximately 1024 m/s.

FAQs:

Q1. What is the difference between mass and weight?

Ans. Mass is a measure of the amount of matter in an object, while weight is a measure of the force of gravity on an object. Mass is measured in kilograms (kg), while weight is measured in newtons (N).

Q2. What is the acceleration due to gravity on Earth?

Ans. The acceleration due to gravity on Earth is approximately 9.8 m/s².

Q3. What is the difference between the gravitational force and the electrostatic force?

Ans. The gravitational force is the force of attraction between any two objects with mass, while the electrostatic force is the force of attraction or repulsion between two charged objects.

Q4. What is the gravitational field?

Ans. The gravitational field is the region around a mass where another mass would experience a force due to gravity. It is a vector field, meaning that at every point in space, there is a gravitational force acting in a certain direction.

Q5. What is escape velocity?

Ans. Escape velocity is the minimum velocity needed to escape the gravitational field of a planet or other celestial body. It is the velocity at which the kinetic energy of an object is equal to its potential energy due to gravity. The formula for escape velocity is given by:

v = sqrt(2GM/r)

Where v is the escape velocity, G is the gravitational constant, M is the mass of the planet or celestial body, and r is the distance from the center of the planet or celestial body.

Q6. What is the difference between gravitational potential energy and gravitational potential?

Ans. Gravitational potential energy is the energy an object possesses due to its position in a gravitational field, while gravitational potential is the potential energy per unit mass at a point in a gravitational field. Gravitational potential is a scalar quantity, while gravitational potential energy is a scalar quantity for a single object, but a vector quantity when considering the potential energy of a system of objects.

Q7. What is the difference between the weight of an object and the force of gravity acting on it?

Ans. The weight of an object is the force with which the object is pulled towards the center of the Earth due to gravity. It is equal to the mass of the object multiplied by the acceleration due to gravity. The force of gravity acting on an object is the force that the Earth exerts on the object due to its mass. It is equal to the product of the mass of the object and the gravitational field strength at the location of the object.

Q8. What is the significance of the gravitational constant?

Ans. The gravitational constant is a fundamental constant of nature that relates the strength of the gravitational force to the masses of the objects and the distance between them. It allows us to calculate the force of gravity between any two objects in the universe. The value of the gravitational constant is crucial in many areas of physics, including astronomy, cosmology, and particle physics.

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GRAVITATION

1.                              In The universe every mass attracts the other mass by a force, which is known as gravitational force.

2.                              Earth also attracts the other masses and the force of attraction of earth is said to be force of gravity.

Newton’s Law of Gravitation: -

This law is for the force of attraction between any two-point masses.

Consider two masses m₁ & m₂ , kept at a distance of  r  in between then the force acting between then will be   ‘F’ which is given as

<![endif]-->                             F                           F

               m₁                                                                  m₂

                                        r

1        The force of attraction is proportional to product of masses say

F  

<![endif]--> m₁ m₂

      2.   The force of attraction is inversely proportional to the square of distance between the masses. i.e.

                                           

3.    The gravitational force is always attractive force.

4.    The gravitational force will be along the line joining the masses.  So,

                          

       G  =  Gravitational Constant having same value in the complete universe.

       G  =    6.67 x 10 ^–11  Nm²     kg ^–2                                                  [G]  =  [M^-1 L³ T^-2]

       F   =    6.67 x 10 ^–11

Definition of unit mass or one kg   :-       

            If   m₁ =  m₂   =   1 kg ,       r  =   1 m

            F = 6.67 x 10^-11  Nm² kg^-2

It two equal masses are kept at a distance of –1m apart and force acting between them is 6.67 x 10^-11 N, then each mass will be unit mass of 1 kg.

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