Sign up

Login
Home

Math

Physics

Chemistry

Biology

Other

Tools
Submit New Formulas/Articles
Recently Added Physical Formulas
·
Electrical Capacitance in an Electronic Circuit
·
Electrical Capacitance
·
Conductor in Static Electric Field
·
Electrical Conductance and Electrical Resistance
·
Current Density
Additional Formulas
·
Coulomb's Law
·
Gauss' Law
·
Application of Gauss’s Theorem
·
Joule's Law
·
Kirchhoff's Current Law
·
Fundamental Postulates of Electrostatics In Free Space
·
Wave
·
Refraction and Reflection
·
Electromotive Force
·
Electrostatic Force
Current Location
>
Physics Formulas
>
Electromagnetism
> Coulomb's Law
Coulomb's Law
According to scientist coulomb, the electric force between two stationary poles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Let us consider two stationary poles have charges q
_{1}
and q
_{2}
and are separated by distance r, then the electrical force (F) will be:
F œ q
_{1}
q
_{2 }
and F œ 1/r
^{2}
F = K· q
_{1}
·
q
_{2}
/ r
^{2}
Where
K = is a constant which has a value of 9 × 10
^{9}
Nm
^{2}
C
^{2}
r = distance in meter (m)
q
_{1,}
q
_{2 }
= charges expressed in coulomb (C)
F = electrical force in Newton (N)
Suppose we consider vacuum as the medium between the two poles, then constant K is replaced by 1/4 π · ε
_{0}
where ε
_{0}
is electrical permittivity of free space.
F = (1/4 π · ε
_{0}
)
_{ }
· (q
_{1}
· q
_{2}
/ r
^{2}
)
Coulomb’s law is used to find the repulsive force acting between two electric charges. In case of more than two electric charges are present, the superposition principle in addition to coulomb’s law is applied to calculate the net electronic force acting on any one charge.
Superposition principle is defined as when more than one coulombian force are acting on a charge, the resultant coulombian force acting on it is equal to the vector sum of the individual forces.
In short, the coulombian force acting between two charges is not affected by the presence of other charge that means coulombian force is a two body force.
According to Coulomb's Law that the magnitude of the electric field (
E
) created by a single point charge (q) at a certain distance (R) is given by:
Where
q
is a point charge.
The electric field intensity of a positive point charge is in the outward radial direction and has a magnitude proportional to the charge and inversely proportional to the square of the distance from the charge. It means if we have an electric field caused by a positive point charge. Then we can measure the strength of that field at different distance, which is
R
. The further we measure the strength, the great become
R
, and the less become the strength of that field at the particular place.
When a charge is not at the center of a coordinate, we can use the following equations:
since
thus:
Calculation:
Example1:
The force acting between two particles is equal to the weight of one of them. The two particles are having same mass 1.6 kg and charge 1.6C. Calculate the distance between them. The value of K = 9 × 10
^{9}
and g = 10 ms
^{2}
.
K = 9 × 10
^{9}
g = 10 ms
^{2}
q
_{1}
= q
_{2}
= 1.6C
m = 1.6 kg
F = mass of one of the particles = mg, which gives:
F = K· q
_{1}
· q
_{2}
/ r
^{2}
= mg
Isolating r from the equation we have:
r
^{2}
= K· q
_{1}
· q
_{2}
/ mg
r
^{2}
= 9 × 10
^{9}
×
(1.6 × 1.6) / (1.6 x 10)
r
^{2}
= 1.440 × 10
^{9}
r = 1.2 × 10
^{3}
Example2:
The force acting between two charges kept at a certain distance is A. Suppose the magnitude of charges is halved and the distance between them is doubled, then the force acting between them is?
a) A/4
b) A/16
c) A/8
d) A
We know that:
F = A
Charges are halved = q
_{1}
/2 q
_{2}
/2
Distance is doubled = 2r
Insert the values into F = K· q
_{1}
· q
_{2}
/ r
^{2}
, we have:
F
_{new}
= K· (q
_{1}
/2) · (q
_{2}
/2) / (2r)
^{2}
F
_{new}
= K· q
_{1}
· q
_{2}
/ (16 r
^{2}
)
And F
_{new}
= F/16, since F = A, we have:
F
_{new}
= A/16
WebFormulas.com ©
2024

Contact us

Terms of Use

Privacy Policy
