Question

Two point charges q and - q are located at points (0 , 0 , - a ) and (0 , 0 , a) respectively :
(a) Find the electrostatic potential at (0 , 0 , z) and ( x, y , 0).
(b) How much work is done in moving a small test charge from the point (5 , 0 , 0 , ) to (-7 , 0 , 0 ) along the x - axis ?
(c) How would your answer change if the path of the test charge between the same points is not along the x axis but along any other random path ?
(d) If the above point charges are now placed in the same position in a uniform external electric field `vecE` , what would be the potential energy of the charge system in its orientation of unstable equilibrium ? Justify your answer in each case .

Answer 1

Two point charges q and - q are located at point (0 , 0 ,a) . And (0 , 0 , z)
(a) Electrostatic potential at (0 , 0 a).
at q (0 ,0 -a)
at -q (0 , 0, a)
`V=(1)/(4piepsilon_(0))((-q)/(z+a))+(1)/(4piepsilon_(0))((q)/(z-a))= (q(z+a)(-z+a))/(4piepsilon_(0)(z^(2)-a^(2)))`
`=(qxx2a)/(4piepsilon_(0)(z^(2)-a^(2)))` " " Dipole moment `[P=q^(xx)2a]`
`=(p)/(4piepsilon_(0)(z^(2)-a^(2)))`
How much work done is moving test charge from ( 5 , 0 , 0 ) to (-7 0 , 0 , ) Work done = qV
Potential at ( 5 , 0 , 0)
`V_(1)=(-q)/(4piepsilon_(0))(1)/(sqrt((5-0)^(2)+(-a)^(2)))+(q)/(4piepsilon_(0))(1)/(sqrt((5-0)^()+a^(2)))=(-q)/(4piepsilon_(0))sqrt(25+a^(2))=0`
Potential at point (-7 , 0 ,0)
`V_(2)=(-q)/(4piepsilon_(0))(1)/(sqrt((-7-0)^(2)+a^(2)))+(q)/(4piepsilon_(0))(1)/((-7-0)^(2)+a^(2))=(-q)/(4piepsilon_(0))*(1)/(sqrt(4a+a^(2)))+(q)/(4piepsilon_(0))(1)/(sqrt(4a+a^(2)))=0` Work done `="Change"xx"Potential"(V=(W)/(Q))`
`"Charge"xx(V_(2)-V_(1))" ""Charge"xx0-0=0 " "W=0`
(c ) As work done does not depend on the path .It depends on the final and initial position . Work done will continue to zero along every path .
(d) Dipole moment of charge system is as p . p `=|q|xx2a` , when dipole is placed external uniform field E is in unstable equilibrium then `theta=180^(@)` then potential energy (P -E) `=-pEcostheta`
`=-qxx2axxEcos180^(@)=qxx2aE`.