Notices
Notice: Exam Form BE IV/II & BAR V/II (Back) for 2076 Magh
Routine: BE IV/II & BAR V/II - 2076 Magh
Result: BCE I/II exam held on 2076 Bhadra
Result: All (except BCE & BEI) I/II exam held on 2076 Bhadra
Notice: Exam Center for Barrier Exam (2076 Poush), 1st Part
View All
View Old Questions
Computer Engineering(BCT)
Electrical Engineering(BEL)
Electronics and Communication(BEX)
View All
View Syllabus
Computer Engineering(BCT)
Electrical Engineering(BEL)
Electronics and Communication(BEX)
View All

Notes of Electromagnetics [EX 503]

Electric Field

 

Coulomb's Law

Coulomb's law states that the force between two very small objects separated by a distance in space is proportional to the product of their charges and inversely proportional to the square of distance between them.
Let Q1 and Q2 be the charges of two small objects separated by distance R; then by Coulomb's law:
F ∝ Q1 . Q2 ..........1
F ∝ 1/R2 .........2

On combining equation 1 and 2,
F ∝ Q1.Q2 / R2
F = k. Q1.Q2 / R2
where, k is a proportionality constant.

The value of k is given by:
k = 1 / 4πεo

Hence, we can write:

  F = (Q1.Q2) / (4πεoR2)
where, 
       εo = 8.854 * 10 -12 C2 kg-1 m3 sec2

 

In vector form:
F2 = (Q1.Q2) / (4πεoR122) . a12
where,
R12 = r2 - r1
a12 = R12 / |R12|

Also,
F1 = - F2

-by SURAJ AWAL

Sponsored Ads