Question
Question: ABC is a right-angled triangle. Calculate the magnitude of the force on a charge –Q. 2=(BA)2+(BC)2
The given value of AC = 5a, BA = 4a
(BC)2=(5a)2−(3a)2
(BC)2=25a2−9a2
BC2=16a2
BC=4a
Now, the force due to the charge at A is given by the above formula,
F1=4π∈1o×9a2(−Q)×(9Q)=4π∈1o×a2(−Q2)
Force on charge C is given as,
F2=4π∈1o×16a2(−Q)×(16Q)=4π∈1o×a2(−Q2)
Therefore total net force is given as,
∣Fnet∣=F1+F2
⇒Fnet=4π∈o1×2a2Q2
⇒Fnet=4π∈o2×r2Q2
Thus, the magnitude of the force on a charge –Q is 4π∈o2×r2Q2
Note: The "Coulomb's law applies to only stationary electric charges" because, in case of moving charges, there is a current, which generates a magnetic field that, in turn, exerts a force on moving charges. Therefore we no longer consider Coulomb's Force alone. Also it signifies the inverse square dependence of electric force. It can also be used to obtain simple derivations of Gauss' law for general cases accurately. The vector form of Coulomb's law is important as it helps us specify electric fields' direction due to charges. It is also written as its abbreviation 'C,' is the SI unit for electric charge. One Coulomb is defined as the amount of charge from a current of one ampere flowing for one second. The charge of an electron is 1.6×10−19C.