The lines of force of the electric field of a positive charge (+q) and a negative charge (-q) are shown in figs. (a) and (b) below. Then :

Options

A. the work done in moving a small positive charge `(+q_(0))` from Q to P will be positive
B. the work done in moving a small negative charge `(-q_(0))` from B to A will be positive
C. in going from Q to P , the kinetic energy of a small negative charge `(-q_(0))` increase
D. in going from B to A , the kinetic energy of a small negative charge `(-q_(0))` decreases .

Answer

B, C, D

Explanation

(b,c,d) In fig (a) , `U = = -q_(0) (V_(Q)-V_(P))`
`( :. W/(q_(0)) = q/(4pi in_(0)r_(Q)) - q/(4piin_(0)r_(p)))`
` :. V(Q) lt V_(p)`
` :. V_(Q) - V_(P) = - ve `
` :. ` Pot energy of charge `(-q_(0))` decreases by some amount . Then K.E increases by same amount .
In fig (b) , `U = -q_(0) (V_(B) - V_(A))`
`V_(B) gt V_(A)`
` :. ` P.E of `-q_(0)` charge increases by some amount . Then K.E decreases by same amount .