express coulombs inverse square law in vector form. how charge is distributed on an irregular conductor
Answers
Answer:
Explanation:
If q is slightly displaced towards A, FA increases in magnitude while FB decreases in magnitude. Now the net force on q is toward A so it will not return to its original position. So for axial displacement, the equilibrium is unstable.
If q is displaced perpendicular to AB, the force FA and FB bring the charge to its original position. So for perpendicular displacement, the equilibrium is stable.
Key Points on Coulomb’s Law
1. If the force between two charges in two different media is the same for different separations, F=\frac{1}{K}\frac{1}{4\pi {{\in }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}F=
K
1
4π∈
0
1
r
2
q
1
q
2
= constant .
2. Kr2 = constant or K1r12 = K2r22
3. If the force between two charges separated by a distance ‘r0’ in a vacuum is the same as the force between the same charges separated by a distance ‘r’ in a medium, then from Coulomb’s Law; Kr2 = r02
4. Two identical conductors having charges q1 and q2 are put to contact and then separated after which each