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Inverse Square Law Formula - Definition, Formula, Solved Questions 

Inverse Square Law Formula: This law states that the intensity or strength of a physical quantity is inversely proportional to the square of the distance from the source.
authorImageMurtaza Mushtaq25 Oct, 2023
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Inverse Square Law Formula

Definition Of Inverse Square Law Formula

Inverse Square Law Formula: The Inverse Square Law is a fundamental principle in physics and mathematics, often used to describe various phenomena, including gravity, electromagnetism, and light. This law states that the intensity or strength of a physical quantity is inversely proportional to the square of the distance from the source. Let's explore this concept in more detail. The Inverse Square Law Formula is a fundamental concept in science and mathematics that plays a significant role in various fields, from physics to engineering and astronomy. It describes how the intensity of a physical quantity diminishes with distance from its source. Understanding this law is essential for a wide range of applications and has shaped our understanding of the physical world.

Formula Of Inverse Square Law

Inverse Square Law Formula: The Inverse Square Law can be expressed mathematically as follows:

I = k / r^2

Where: - I represents the intensity of the effect or force. - k is a constant that depends on the specific situation. - r is the distance from the source. Here are some of the formulas used in the context of the Inverse Square Law:
  1. Newton's Law of Universal Gravitation:
- F = G * (m1 * m2) / r^2, where F is the gravitational force between two objects, G is the gravitational constant, m1 and m2 are the masses of the objects, and r is the distance between them.
  1. Coulomb's Law:
- F = k * (q1 * q2) / r^2, where F is the electrostatic force between two charges, k is Coulomb's constant, q1 and q2 are the charges, and r is the separation between them. This formula is used in a wide range of fields, so let's delve into some of the key applications:
  1. Gravity:
- The law of universal gravitation, formulated by Isaac Newton, follows the Inverse Square Law. It explains how the force of gravity weakens as you move farther from a massive object.
  1. Electromagnetism:
- Coulomb's Law, which describes the force between charged particles, also adheres to the Inverse Square Law. It is crucial in understanding electrical and magnetic interactions.
  1. Light and Radiation:
- In optics, the intensity of light follows this law. It's why objects appear dimmer as they move farther from a light source.
  1. Sound:
- The intensity of sound waves spreading from a point source decreases with distance, adhering to the Inverse Square Law.
  1. Radioactive Decay:
- In nuclear physics, the law governs the rate at which radioactive materials decay over time.
  1. Acoustics:
- The law is essential in designing sound systems and understanding sound propagation in open spaces.

Solved Examples Of Inverse Square Law Formula

Numerical Problem 1: Suppose you have two point charges, q1 = 4 μC and q2 = 8 μC, placed 3 meters apart in a vacuum. Calculate the electrostatic force between them. Solution:
  1. We'll use Coulomb's Law: F = k * (q1 * q2) / r^2
  2. Plug in the values:
- F = (9 x 10^9 Nm^2/C^2) * (4 x 10^(-6) C) * (8 x 10^(-6) C) / (3 m)^2
  1. Calculate the result:
- F = (9 x 10^9) * (32 x 10^(-12)) / 9 - F = 32 x 10^(-3) N - F = 0.032 N So, the electrostatic force between the two charges is 0.032 Newtons. Numerical Problem 2: A light source emits 1600 lumens of light. If you move 2 meters away from the source, what will be the illuminance at that distance? Solution:
  1. We'll use the Inverse Square Law for light intensity: I = L / (4πr^2)
  2. Plug in the values:
- I = 1600 lumens / (4π * (2 m)^2)
  1. Calculate the result:
- I = 1600 / (4π * 4) - I ≈ 31.83 lumens/m^2 So, the illuminance at a distance of 2 meters from the light source is approximately 31.83 lumens per square meter. Numerical Problem 3: A radioactive sample initially contains 2000 radioactive atoms. After 10 hours, only 500 atoms remain. Calculate the decay constant for this sample. Solution:
  1. We'll use the decay formula N(t) = N₀ * e^(-λt), where N(t) is the number of atoms at time t, N₀ is the initial number of atoms, λ is the decay constant, and t is time.
  2. Plug in the given values: N(t) = 500, N₀ = 2000, and t = 10 hours.
  3. Solve for λ:
- 500 = 2000 * e^(-10λ)
  1. Divide both sides by 2000:
- 0.25 = e^(-10λ)
  1. Take the natural logarithm of both sides to solve for λ:
- ln(0.25) = -10λ
  1. Divide by -10 to find λ:
- λ ≈ -0.1155 per hour So, the decay constant for this radioactive sample is approximately -0.1155 per hour. These are three numerical examples related to the Inverse Square Law, Coulomb's Law, and radioactive decay. If you have more questions or need additional examples, please feel free to ask. Certainly! Here are 10 more numerical problems along with their step-by-step solutions, covering various topics related to the Inverse Square Law, physics, and mathematics: Numerical Problem 4: Suppose you have two masses, m1 = 5 kg and m2 = 8 kg, placed 2 meters apart. Calculate the gravitational force between them. Solution:
  1. We'll use Newton's Law of Universal Gravitation: F = G * (m1 * m2) / r^2.
  2. Plug in the values:
- F = (6.67 x 10^(-11) Nm^2/kg^2) * (5 kg) * (8 kg) / (2 m)^2.
  1. Calculate the result:
- F = (6.67 x 10^(-11)) * (40) / 4. - F = 6.67 x 10^(-10) N. The gravitational force between the two masses is 6.67 x 10^(-10) Newtons. Numerical Problem 5: A radioactive sample has an initial count of 1000 disintegrations per minute (DPM). After 30 minutes, it has decreased to 125 DPM. Calculate the decay constant. Solution:
  1. We'll use the decay formula N(t) = N₀ * e^(-λt).
  2. Plug in the values: N(t) = 125 DPM, N₀ = 1000 DPM, and t = 30 minutes (0.5 hours).
  3. Solve for λ:
- 125 = 1000 * e^(-0.5λ).
  1. Divide both sides by 1000:
- 0.125 = e^(-0.5λ).
  1. Take the natural logarithm of both sides to solve for λ:
- ln(0.125) = -0.5λ.
  1. Divide by -0.5 to find λ:
- λ ≈ 2.772 per hour. The decay constant for this radioactive sample is approximately 2.772 per hour. Numerical Problem 6: You have a light source emitting 8000 lumens. What will be the illuminance at a distance of 4 meters from the source? Solution:
  1. We'll use the Inverse Square Law for light intensity: I = L / (4πr^2).
  2. Plug in the values: I = 8000 lumens and r = 4 meters.
  3. Calculate the result:
- I = 8000 / (4π * (4 m)^2). - I ≈ 12.73 lumens/m^2. The illuminance at a distance of 4 meters from the light source is approximately 12.73 lumens per square meter. Numerical Problem 7: You're standing 5 meters from a sound source that emits 80 dB of sound. What will be the sound intensity at your location? Solution:
  1. We'll use the formula for sound intensity in decibels (dB): I = I₀ * 10^(dB/10).
  2. Plug in the values: I₀ = 10^(-12) W/m^2 (threshold of hearing) and dB = 80 dB.
  3. Calculate the result:
- I = (10^(-12) W/m^2) * 10^(80/10). - I = 10^(-4) W/m^2. The sound intensity at your location is 10^(-4) watts per square meter. Numerical Problem 8: You have a radioactive sample with an initial count of 2000 disintegrations per second (DPS). After 5 seconds, it has decreased to 1000 DPS. Calculate the decay constant. Solution:
  1. We'll use the decay formula N(t) = N₀ * e^(-λt).
  2. Plug in the values: N(t) = 1000 DPS, N₀ = 2000 DPS, and t = 5 seconds.
  3. Solve for λ:
- 1000 = 2000 * e^(-5λ).
  1. Divide both sides by 2000:
- 0.5 = e^(-5λ).
  1. Take the natural logarithm of both sides to solve for λ:
- ln(0.5) = -5λ.
  1. Divide by -5 to find λ:
- λ ≈ 0.1386 per second. The decay constant for this radioactive sample is approximately 0.1386 per second. Numerical Problem 9: A satellite is orbiting the Earth at a distance of 7000 kilometers from the center of the Earth. Calculate the gravitational force acting on the satellite if its mass is 500 kg. Solution:
  1. We'll use Newton's Law of Universal Gravitation: F = G * (m1 * m2) / r^2.
  2. Convert the distance to meters: 7000 km = 7,000,000 meters.
  3. Plug in the values:
- F = (6.67 x 10^(-11) Nm^2/kg^2) * (500 kg) * (5.972 x 10^24 kg) / (7,000,000 m)^2.
  1. Calculate the result:
- F = (6.67 x 10^(-11)) * (2.986 x 10^27) / 49,000,000,000 m^2. - F ≈ 1.283 x 10^3 N. The gravitational force acting on the satellite is approximately 1,283 Newtons. Numerical Problem 10: You're conducting an experiment with radioactive decay. A sample has an initial count of 300 disintegrations per second (DPS). After 20 seconds, it has decreased to 75 DPS. Calculate the decay constant. Solution:
  1. We'll use the decay formula N(t) = N₀ * e^(-λt).
  2. Plug in the values: N(t) = 75 DPS, N₀ = 300 DPS, and t = 20 seconds.
  3. Solve for λ:
- 75 = 300 * e^(-20λ).
  1. Divide both sides by 300:
- 0.25 = e^(-20λ).
  1. Take the natural logarithm of both sides to solve for λ:
- ln(0.25) = -20λ.
  1. Divide by -20 to find λ:
- λ ≈ 0.0922 per second. The decay constant for this radioactive sample is approximately 0.0922 per second. The Inverse Square Law Formula is a fundamental concept in science and mathematics that plays a significant role in various fields, from physics to engineering and astronomy. It describes how the intensity of a physical quantity diminishes with distance from its source. Understanding this law is essential for a wide range of applications and has shaped our understanding of the physical world.

Inverse Square Law Formula FAQs

Why is it called the "Inverse Square Law"?

It's named this way because the strength or intensity of the effect decreases with the square of the distance from the source.

What happens if the distance (r) is doubled?

If the distance is doubled, the intensity or force becomes one-fourth of its original value because (2^2 = 4).

What are some real-world applications of the Inverse Square Law?

Real-world applications include satellite orbits, GPS systems, radiation therapy, and architectural lighting design.

How does the Inverse Square Law apply to photography?

It's crucial in understanding how light intensity falls off with distance from a light source and is essential in achieving proper exposure in photography.
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