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 the wrong option is

Easy

A

the work done in moving a small positive charge (+q₀) from Q to P will be positive
B

the work done in moving a small negative charge (-q_o) from B to A will be positive
C

in going from Q to P the kinetic energy of a small negative charge (-q_o) increases
D

in going from B to A, the kinetic energy of a small negative charge (- q_o) decreases.

Solution

In fig. $(a); U = - q_{0} (V_{Q} - V_{P})$
$(∵ \frac{W}{q_{0}} = \frac{q}{4 {πϵ}_{0} r_{Q}} - \frac{q}{4 {πϵ}_{0} r_{P}})$
$\begin{array}{l} ∵ V_{Q} < V_{P} \\ ∴ V_{Q} - V_{P} = - {ve}^{} \end{array}$
$∴$ Pot energy of charge (- q_o) decreases by some amount. Then K.E. increases by same amount. In fig. (b); U=-q_o (V_B- V_A) V_B>V_A
$∴$ PE. of-q_o charge increases by some amount. Then K.E. decreases by same amount.

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