Figures 2.8 (a) and (b) show the field lines of a positive and negative point charge respectively.

(a) Give the signs of the potential difference `V_(P)-V_(Q): V_(B)-V_(A)`.
(b) Give the sign of the potential energy difference of a small negative charge between the points Q and P, A and B.
(c) Give the sign of the work done by the field in moving a small positive charge from Q to P.
(d) Give the sign of the work done by the external agency in moving a small negative charge from B to A.
(e) Does the kinetic energy of a small negative charge increase or decrease in going from B to A?

Explanation

In V`=(k(+q))/(r) ` k(+q) is constant
`:. V prop (1)/(r)`
Let the distance of point P from + q charge is `r_(p)` and distance of point Q is `r_(Q)`,
`r_(P) lt r_(Q)`
`:. V_(P) gt V_(Q) implies V_(P)-V_(Q) gt 0 :. ` Positive
and in `V=(k(-q))/(r)` if kq is considered as positive
`V prop -(1)/(r)`
Let the distance of point A from - q chargd is .A and distance of point B is `r_(B)`
`r_(A) lt r_(B)`
`:. V_(A) -V_(B) lt 0`
`:. V_(B) - V_(a) gt 0`
(b) Potential energy decrease as attraction force acts. Here attraction force acts when negative charge is placed in the field of positive charge and hence potential energy decreases so,
`U_(Q) gt U_(P)`
`:. U_(Q) - U_(P) gt 0 " " :.` Positive and when negative charge is placed in the field of negative charge and hence potential energy increases so `U_(A) lt V_(B)` taking from B to A
`:. U_(B) -U_(A) gt 0 " " :.` Positive
( c) In moving a positive charge from Q to P, work has to be done against the electric field. Therefore, work done by the field is negative but work done on the field is positive.
(d) In moving a negative charge from B to Aworlc has to be don.e by the external agency. It is positive.
(e) Due to force of repulsion on the negative charge, velocity decreases and hence the lcinetic energy decreases in going from B to A.