Coulomb’s Law – Has Sinaumed’s ever seen a phenomenon when we bring the ends of a battery closer but cannot attach the two ends?
If so, Sinaumed’s is directly observing the phenomena of electrostatics. This phenomenon can occur because the tip of the battery has a positive (+) and negative (-) electric charge. When we bring the end that has a positive charge closer to the end that also has a positive charge, the two batteries seem to refuse to stick. However, if the end that has a positive charge is brought close to the end that has a negative charge, it will easily stick together.
This electrostatic phenomenon will then be discussed in more depth through a law called Coulomb’s law.
History of Coulomb’s Law
Coulomb’s law was first put forward by a French physicist, namely Charles Augustin de Coulomb in the 1780s. At first the development of this law actually started in 600 BC. It all started with the discovery of the theory of static electricity. Then, the theory continues to be developed by other scientists.
In 1600, a British scientist named William Gilbert conducted a study on static electricity which was used to distinguish it from the effects of magnetic stones. From William Gilbert’s research, the word electric was found in electricity terms.
Then in the 18th century, two scientists from Italy, namely Daniel Bernoulli and Alessandro Volta conducted an experiment to measure the force between the plates of a capacitor to find out whether the electric force decreases with distance and is affected by the force of gravity. Besides that, Franz Aepinus also discovered the inverse square law in 1758.
After that there were also experiments conducted by scientists from England, namely Joseph Priestley. Joseph Priestley conducted experiments on balls that had an electric charge. He proposed that the electric force followed an inverse square law, but he did not go into detail and simply assumed that the force between charges varies as the inverse square of the distance.
Then, the climax was research conducted by Charles Augustin de Coulomb in 1785. In this research, Coulomb conducted experiments using a torsion balance which is useful for knowing the magnitude of the force acting on two objects that have an electric charge.
As a result, Coulomb got the result, namely the theory of static electricity which was eventually named Coulomb’s Law. Apart from being enshrined as a legal name, the name Coulomb is also enshrined as a unit of electric charge, namely Coulomb (C).
Definition of Coulomb’s Law
Coulomb’s law is basically a law that explains the relationships that exist in electric charges. As is well known, that electric charge is divided into two types, namely positive charge and negative charge.
This law describes the condition when two electric charges with a certain distance interact with each other and exert an attractive or repulsive force. One of the factors that influence the magnitude of the Coulomb’s law force itself is the magnitude of the object’s electric charge.
An attractive force will occur when dissimilar electric charges (+-) meet, while a repulsive force will occur when similar electric charges (++/–) meet each other.
Coulomb’s law actually has similarities with the leading law of gravity that has been put forward by Isaac Newton. The equation lies in the inverse square ratio. While the difference is the force produced, where the gravitational force only occurs attraction, while the electric force is not only an attractive force but also a repulsive force.
Coulomb’s law sounds
Based on research conducted by Charles Auguste de Coulomb. So the conclusion appears that the sound of Coulomb’s Law is:
“If there are two electrically charged objects it will cause a force between them, namely attraction or repulsion, the magnitude will be directly proportional to the product of the values of the two charges and inversely proportional to the square of the distance between the two objects.”
As previously explained, when two similar charges are brought closer they will cause mutual repulsion, whereas if there are two different charges if they are brought closer they will cause mutual attraction and can stick. These two forces are called electrostatic forces.
Application in Everyday Life
A simple example that Sinaumed’s can see in everyday life is the effect that occurs when two ends of a magnet are brought closer to each other. If the side of the magnet that has a negative charge is brought close to the side of another magnet that has a positive charge, or iron which has a positive charge, then the magnet will stick, and vice versa if the side of the magnet that is negatively charged is brought close to the side of the magnet which is also negatively charged, rejection will occur. .
Another example is the application of lightning rods which are usually in various homes. The workings of lightning rods generally have a positive charge. So that when there is lightning that has a negative charge, the lightning rod will catch the lightning and then flow it to the ground using a conductor cable and make the lightning flow not enter into the electricity that is in the house.
Coulomb’s Law Formula
From the statement expressed by Charles Auguste de Coulomb, the mathematical formula used to determine the magnitude of the Coulomb force appears, which is as follows:
Information:
F : Gaya Coulomb (N)
K : Coulomb constant ( 9×109 Nm2/C2)
q1 : Mass of object 1 (C)
q2 : Amount of charge of object 2 (C)
r : Distance between charges (m)
Coulomb (C) unit size
1 micro coulomb (µC) = 1×10-6 C
1 nano coulomb (nC) = 1×10-9 C
1 coulomb = 1 coulomb (C)
From this formula it can be concluded that the coulomb constant value will depend on the unit selected.
Thus, it can be interpreted that the farther the distance between the two charges will cause the electric force that is also smaller. Conversely, if the distance between the two charges is close, it will cause the electric force to become even greater.
Problems example
Example question number 1
It is known that two magnets each have an electric charge of 4 C and 2 C and the two magnets are 3 meters apart. Then determine how many coulomb forces occur from the two charges!
Answer:
Is known:
F : ?
K : 9×109 Nm2/C2
q1 : 4 C
q2 : 2 C
r : 3 meter
Then the coulomb force resulting from the two electric charges is 8 × 109 N.
Two pieces of iron each have an electric charge of 4 × 10-6 C and 6 × 10-6 C and have a distance of 2 cm. Determine how many coulomb forces are generated by the two irons!
Answer:
F : ?
K : 9×109 Nm2/C2
q1 : 4×10-6 C
q2 : 6×10-6 C
r : 2 cm = 0,02 m
109
So the coulomb force found on the two objects is 540 N.
Example question number 2
Two pieces of iron have electric charges of 10 micro coulombs (µC) and 50 micro coulombs (µC) respectively. When the distance between the two charged irons is 10 cm. How much coulomb force arises from the two electrically charged irons?
Answer:
F : ?
K : 9×109 Nm2/C2
q1 : 10 µC = 10×10-6 C
q2 : 50 µç = 50×10-6 C
r : 10 cm = 0,1 m
Then the coulomb force generated from the two irons is 450 N.
Example question number 3
Two objects have charges of 3×10 -6 C and 6 x 10 -6 C. The distance between them is 3 cm. How big is the electric force on each charge?
Is known
Q1 = 3×10-6 C
Q2 = 6 x 10-6C
r = 3 cm = 3 x 10-2 m
k = 9 x 109 Nm2/C2
Solution
Fc = k(q1q2/r2)
Fc = 9 x 109 (3×10-6 C x 6 x 10-6C)/9 x 10-4
Fc = 9 x 109 ( 18 x 10-12)/ 9 x 10-4
Fc = 1,8 x 102 N
Example question number 4
When 2 like charges have a charge size of 5 x 10 -4 C and 5 x10 -4 C. The 2 charges are separated by a distance of 5 cm. then how big is the coulomb force generated? (k = 9 x 10 9 Nm 2 /C 2 )
Is known
Q1 = 5 x 10-4C
Q2 = 5 x 10-4C
r = 5 cm = 5 x 10-2 m
k = 9 x 109 Nm2/C2
Solution
Fc = k(q1q2/r2)
Fc = 9 x 109 (5 x 10-4C x 5 x 10-4C)/25 x 10-4
Fc = 9 x 109 (25 x 110-8)/ 25 x 10-4
Fc = 9 x 105 N
Example question number 5
Look at the picture below. The magnitude of the electric force acting on the charge is 27 Newton. If it is known that the charge at B is +6 micro Coulomb, calculate how many micro Coulomb the charge is at point C.( k = 9 x 109N m2/C2 and 1 micro Coulomb = 10−6 C)?
Discussion
Is known:
q1= Charge at Point B = +6 µC = +6×10−6 Coulomb
k = 9 x 109N m2/C2
r = 10 cm = 0,1 m
F=27 Newtons
Asked:
q2= Charge at Point C
Answer:
For:
27 = (9🗙 109 ) 🗙 (6×10−6) 🗙 (q2 🗙10−6 )
______________________________
(0,1)²
(q2) =27 🗙 (0,1)²
__________
(9 🗙109) 🗙 (6🗙10-12)
(q2 )=Charge at Point C= – 5 micro Coulomb
The charge at Point C is – 5 micro Coulombs
Example question number 6
The charge on object A is +8 micro Coulomb and the charge on object B is -10 micro Coulomb. The magnitude of the electric force acting on the two charges is 18 Newton if it is known that 9 x 109N m2/C2 and 1 micro Coulomb = 10−6 C, calculate the distance between the charges!
Discussion
Is known:
q1= Charge at Point A = +8 µC = +8×10−6 Coulomb
q2= Charge at Point B = -10 µC = -10×10−6 Coulomb
k = 9 x 109N m2/C2
F= 18 Newtons
Asked:
Distance between loads (r) ?
Answer:
18= (9 🗙 109 ) 🗙 (8🗙10−6 ) 🗙 (10🗙10−6 )
______________________________
r²
r²=(9 🗙 109 ) 🗙 (8🗙10−6 ) 🗙 (10🗙10−6 )
______________________________
18
r²=0,04
r= 0,2 m= 20 cm
So, the distance between the charges is 20 cm
Example question number 7
It is known that two fruits have an electric charge A = +15 micro Coulomb and an electric charge B = +10 micro coloumb separated by a distance as shown below. Determine the magnitude of the coloumb force acting on the two charges?( k = 9 x 109N m2/C2 and 1 micro Coulomb = 10−6 C)
Discussion
Is known:
q1= Charge at Point A = +15 µC = +15×10−6 Coulomb
q2= Charge at Point B = +10 µC = +10×10−6 Coulomb
k = 9 x 109N m2/C2
r = 10 cm = 0,1 m
Asked:
The magnitude of the electric force (F) acting on the charge
Answer:
F= (9 🗙 109) 🗙 (15🗙10−6 ) 🗙 (10🗙10−6 )
______________________________
0,1²
F= 135 Newtons
The magnitude of the electric force (F) acting on the charge is 135 Newton
Example question number 8
It is known that two electric charges E = -10 micro Coulomb and an electric charge F = -15 micro coloumb are separated by a distance as shown below. Determine the magnitude of the coloumb force acting on the two charges (k = 9 x 109N m2/C2 and 1 micro Coulomb = 10−6 C)
Discussion
Is known:
q1= Charge at Point E = -10 µC = +10×10−6 Coulomb
q2= Charge at Point F = -15 µC = -15×10−6 Coulomb
k = 9 x 109N m2/C2
r = 10 cm = 0,1 m
Asked:
The magnitude of the electric force (F) acting on the charge
Answer:
F= (9 🗙 109 ) 🗙 (10🗙10−6 ) 🗙 (15🗙10−6 )
______________________________
0,1²
F= 135 Newtons
The magnitude of the electric force (F) acting on the charge is 135 Newton
Example question number 9
It is known that there are three electric charges placed separately as shown! The magnitude of the charge P is = -4 micro Coulomb, the magnitude of the charge Q is = +10 micro Coulomb and the charge R = -12 micro Coulomb. Determine the magnitude and direction of the electric force on the charge Q when ( k = 9 x 109N m2/C2 and 1 micro Coulomb = 10−6 C) ?
Discussion
Is known:
qP= Charge at Point P = -4 µC = -4× 10−6 Coulomb
qQ= Charge at Point Q = +10 µC = +10× 10−6 Coulomb
qR= Charge at Point R = -12 µC = -12× 10−6 Coulomb
k = 9 x 109N m2/C2
rPQ = 6 cm = 0,06 m
rQR = 4 cm = 0,04 m
Asked:
magnitude and direction of the electric force on the charge Q
Answer:
It should be remembered that the electric force on the charge Q is the resultant electric force between the charges P and Q with the electric force between the charges Q and R
The magnitude of the electric force (FPQ) of the P and Q charges:
F= (9 🗙 109 ) 🗙 (4🗙10−6 ) 🗙 (10🗙10−6 )
______________________________
0,06²
F= 100 Newtons
(The direction of charge goes out from + and in to -)
So, the charge P is negative and the charge Q is positive so that the direction of the Coulomb force is towards the charge P and away from the charge Q (to the left).
The magnitude of the electric force (FQR) of the charges Q and R:
F= (9 🗙 109 ) 🗙 (10🗙10−6 ) 🗙 (12🗙10−6 )
______________________________
0,04²
F= 675 Newtons
(The direction of charge goes out from + and in to -)
The charge Q is positive and the charge R is negative so that the direction of the Coulomb force is towards the charge R and away from the charge Q (to the right).
The resultant electric force on the charge Q :
The FPQ direction is to the left and the FQR direction is to the right.
FQ= FQR– FPQ = 675 N – 100 N = 575 Newtons.
The direction of the resultant electric force on the charge Q (FQ) = the direction of the electric force FQR, namely towards the charge R (to the right) with a value of 575 Newton