Suppose that the same system of charge is now placed in an external electric field E = A(l/`r^(2)), A= 9 xx 10^(5) cm^(-2)`. What would the electrostatic energy of the configuration be ?

Explanation

Total electrostatic energy of two charges in an external electric field,
`U= q_(1)V(r_(1))+q_(2)V(r_(2))+(kq_(1)q_(2))/(r_(12))`
Here V = Er
`=(A)/(r^(2))xxr`
`V = (A)/(r)`
`:. V(r_(1))=(A)/(r_(1)) ` and `V(r_(2))=(A)/(r_(2))`
and `r_(12) = r = 0.18 ` m
`q_(1)= 7 xx10^(-6) C, q_(2)=-2xx10^(-6)C`
` A = 9xx10^(5) NC^(-1) m^(2)`
`:.` From equation (2)
`U =7xx10^(6)(A)/(r_(1))+(-2xx10^(-6))xx(A)/(r_(2))+(kq_(1)q_(2))/(r_(12))`
`:. U=9xx10^(+5)xx(7xx10^(-6))/(0.09)`
`-(9xx10^(+5)xx2xx10^(-6))/(0.09)-0.7`
`[ because` From equation (1)]
=70 -20 -0.7
`:. U = 49.3` J