A system has two charges `q_(A)=2.5xx10^(-7) C` and `q_(B)=-2.5xx10^(-7) C` located at points A, `(0, 0, -0.15 m)` and B, `(0, 0, +0.15 m)` respectively. What is the total charge and electric dipole moment of the system?
Net charge `=2.5xx10^(-7) -2.5xx10^(-7)=0`
Electric dipole moment, `P = ("Magnitude of charge")xx("Separation between charges")`
`=2.5xx10^(-7) [0.15+0.15]C m =7.5xx10^(-8) Cm`
the direction of dipole moment is from B to A.
(i). Here, in the figure, the electric lines offorce emanate from A and C. Therefore, charges A and C must be positive.
(ii) The number of electric lines of forces...
`DeltaU=U_(f)-U_(i)-0=U_(r)`
We have to simply calculate the electrostatic potential energy of the given system of charges
`:. DeltaU=U_(f)=1/(4pi epsi_(0)) (q_(1)q_(2))/r=(9xx10^(9)xx2xx10^(-5) xx2xx10^(-5)xx100)/(10) J=36 J`
`:.` Work required `=36 J =` equal...
Correct Answer - (a) Radius 4a, cantre at `(5a,0)`
(b). `V=(Q)/(4piepsilon_(0))[(1)/(|(x-3a)|)-(2)/(|(x+3a)|)]`
(c). At `x=9a` where `V=0`, the charged particle eventually crosses the circle
`v=sqrt((Qq)/(8piepsilon_(0)ma))`