Difference between Coulomb Force and Gravitational Force
Last Updated :
04 Feb, 2024
In physics, a force is an effect that can alter an object’s motion. A force can cause an object with mass to accelerate by changing its velocity (e.g., moving from a state of rest). Intuitively, the force can be described as a push or a pull. A force is a vector quantity since it has both magnitude and direction. The SI unit of newton is used to measure it (N). The letter F is used to indicate force.
Coulomb’s Force
The force of attraction or repulsion between two electric charges acting along a straight line is proportional to the product of the charges and inversely proportional to the square of the distance between them.
F ∝ q1q2
or, F ∝ 1/r2
F = k × q1 × q2/r2
Where,
q1 and q2 are two charges
F is the force of attraction/repulsion between them
r is the distance between the charges
k is proportionality constant and equals to 1/4 π ε0 = 9 × 109 Nm2/ C2
ε0 is epsilon naught
Gravitational Force
The gravitational force is a force that attracts all mass-bearing objects. The gravitational force is referred to as attractive because it always strives to pull masses together rather than push them apart. Newton’s Universal Law of Gravitation is the name for this. Objects that are really far apart do not tug on each other in a noticeable way. However, the force exists and can be calculated.
Universal Gravitation Equation
F = G × M × m/r2
In the equation:
F is the force of gravity (measured in Newtons, N)
G is the gravitational constant of the universe
M is the mass of one object (measured in kilograms, kg)
m is the mass of the other object (measured in kilograms, kg)
r is the distance those objects are apart (measured in meters, m)
Difference between Coulomb’s Force and Gravitational Force
The main distinction between gravitational and electrostatic forces is that gravitational force is the force that causes the earth to attract other things due to its mass. It is a conservative political party. Coulomb’s force is the force exerted by an object as a result of its charge. It is a conservative force as well.
| Sr. No. |
Gravitational Force
|
Coulomb’s Force
|
| 1. |
It is the force exerted by an object as a result of its mass. |
It is the force exerted by an object as a result of its charge. |
| 2. |
It is a very strong force as compared to the electric force. |
It is a very weak force as compared to the Electric force. |
| 3. |
It is only attractive. |
It is attractive as well as repulsive. |
| 4. |
It does not depend on the medium. |
It depends on the medium. |
| 5. |
Its constant is “G”. |
Its constant is “K”. |
| 6. |
It originates from the mass. |
It originates from the charge. |
| 7. |
Its formula is:
F = G × M × m/r2
|
Its formula is:
F = k × q1 × q2/r2
|
Sample Problems
Problem 1: What is the amount of the force exerted by a 25 μC charge 8.5cm away from a 10 μC charge?
Solution:
Applying Coulomb’s law, we can find the magnitude of the electric force :
F = k× q1×q2 / r2
= (9×10 9)× (25×10 −6C)×(10×10 −6C)/(8.5×10 −2m) 2
= 311.5 N
The two point charges have opposite signs, so the electrostatic force between them is attractive.
Problem 2: At a distance of d=5cm, two like and equal charges produce a force of F=9×103 N on each other. How do you determine the amount of each charge?
Solution:
Coulomb’s law determines the magnitude of the force between two rest point charges q and q’ separated by a distance d.
F = k ×q×q’/d2
9×103 = (8.99×109)×q2/(0.05)2
q2 = 25 ×10-16
q = 5×10-8 C
Problem 3: Calculate the gravitational pull between a massless entity and the Moon. (The moon’s mass is 7.4 x 1022 kg, and its radius is 1.74 x 106 m.)
Solution:
Given,
The mass of the body m=1 Kg
The mass of the moon M=7.4 x 1022 kg
Radius of the moon R=1.74 x 106 m
F = G x M x m/r2
= (6.67×10-11 x1 x 7.4×1022)/ (1.74 x 106)2
=1.63 N
Problem 4: Calculate the gravitational attraction between two protons separated by one angstrom. (proton mass = 1.6 x 10-27 kg)
Solution:
Given,
The mass of protons Mp=1.6×10-27 kg
Distance r= 1 Å = 10-10 m
We know that,
F = G x M x m/r202
= 6.67×10-11 x (1.6×10-27)2/ (10-10)2
= 17 x 10-45 N
Problem 5: A q=4μC point charge is 3cm apart from a q’=1μC charge. Calculate the magnitude of the Coulomb force exerted by one particle on the other.
Solution:
Given,
q = 4μC
q’ = 1μC
d = 3 cm = 3×10-2 m
Coulomb’s law determines the size of the electric force between two stationary point charges,
F = kxqxq’/r2
= (8.99×109)×(4×10−6)(1×10-6)/(0.03)2
= 40 N
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