Electrostatic Force Vs Gravitational Force

Force is merely divers as a push or pull of an object with the mass that causes it to change its velocity. It is an external agent that is capable of changing the state of a item torso neither in a rest position nor in motion. It has both magnitude and direction. It tin be measured using a spring residuum. The formula of force is F = ma, where g = mass of the object, a = acceleration.

Both Gravitational force and Electrostatic strength are examples of Not-Contact forces. Non-Contact forces are defined every bit the force that acts through spaces without making any directly contact with the body.

Gravitational Forcefulness

Gravitational Force is defined as the strength by which the earth attracts some other object by mass. It is the force of attraction between all the masses in the universe, i.e. the attraction of the earth's mass to bodies near its surface. Gravitational Force is the forcefulness that attracts any two objects with the mass

Formula of Gravitational Strength

F = (1000 m_onem₂)/rⁱⁱ

Where,

One thousand = Gravitational constant = six.67 × 10^-11 N. m²/kg²
yard_i= mass of the object 1.
m_two = mass of the object ii.
r = distance between two objects.
South.I units = Newton (N).

Electrostatic Force

It is defined as a force between objects due to the charge of the objects. It is the force exerted by a charged trunk on another charged body or an uncharged trunk. It is the force that comes when the bodies are not in contact. Information technology is also defined as the attractive or repulsive force betwixt two objects that are caused past their electric charge. It is also called Coulomb Force.

Formula of Electrostatic Strength

F = (kq₁q₂)/r²

Where,

k = proportionality constant = 1/4π€_o

Where €_o= permittivity of a vacuum= 9 × 109 N.m²/c².

q_ane= charge of the object 1.

q₂ = charge of the object two.

r = distance betwixt the 2 objects.

Difference between Gravitational Force and Electrostatic Force

	Gravitational Forcefulness	Electrostatic Forcefulness
1.	It is the forcefulness by which earth attracts another object past its mass.	Information technology is the force of an object due to its charge.
2.	The force due to gravity does not depend on the medium.	It depends on the medium (charge).
3.	It is the only bonny force.	Information technology is both attractive as well equally a repulsive force.
4.	Formula: F = (G grand₁m₂)/r²	Formula: F = (kq₁q₂)/r²
5.	It has a gravitational constant (G).	Information technology has a proportionality abiding (k).
6.	It originates from the mass of the object.	It originates from the charge of the object.

Sample Questions

Question 1: How much force is required to accelerate a 2000kg car at 5m/s²?

Answer:

Given that,

mass of the car (m) = 2000kgs.

acceleration (a) = 5 yard/s².

F = ma

F = 2000 × 5

F = 10000 Northward.

Therefore, the force required is 10000 Northward.

Question 2: 2 objects of mass 1000kg and 500 kg are kept at a altitude of 400. Find the force of attraction between them

Answer:

Given that,

m_i= 1000kg

k_ii= 500m

r = 400m

F = Thousand m₁k₂/r²

= (6.67 × 10^-11) × (k × 500)/(400)^two

= (6.67 × 10^-11) × fifty/16

= 2.08 × 10^-10 N.

Therefore, the force of attraction between them is ii.08 x ten^-x N.

Question 3: Two charges 3C and vi C are kept at a altitude of 9m away from each other. Find the force of attraction betwixt them?

Answer:

Given that,

q₁= 3C

q₂= 6C

r = 9m

F = k.q₁q_ii/r²

F = (9 × 10^ix × 3 × 6)/9²

F= ii × 10^ix N.

Therefore, the force of attraction between them is 2 × 10^ix Northward.

Question 4: Permit u.s.a. suppose 2 satellites are passing around the orbit, globe shut to each other. For a moment they are part of each other if 400m. The masses of the satellite are 400kg and 1200kg. Summate the gravitational force between them?

Answer:

Given that,

m₁= 400kg

thousand_two= 1200kg

r = 400m

F = (half-dozen.67 × ten^-11 × 400 × 1200)/(400)²

F = 20.01 × 10^-11 N.

Therefore, the gravitational force betwixt is 20.01 × 10^-xiN.

Question 5: A body of charges 10 C and 30C are kept at a distance of 10m away from each other. Find the electrostatic force between them?

Answer:

Given that,

q₁ = 10C

q₂= 30C

r = 10m

F = (9 × 10⁹ × 10 × xxx)/(10)²

F = 27 × 10⁹ N.

Therefore, the electrostatic force betwixt them is 27 × 10⁹ N.

Question vi: An object having a mass of 2 kg is moving with a speed of 8 m/s hits a wall and comes to rest for 10sec. Calculate the force that makes the object stop.

Respond:

Given that,

chiliad = two kg

Initial velocity u = 8 m/s

Final velocity v = 0 m/due south

Time = 10 sec.

Dispatch a = ( 5 – u)t

a = (0 – viii)10

a = eighty

F = two × 80

F = 160 N.

Therefore, the force that is needed to stop the object = 160 Due north.