Question

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

(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 (−$q_0$) increases
(d) In going from B to A, the kinetic energy of a small negative charge (−$q_0$) decreases

• A. Only a
• B. a and b
• C. a, c, d
• D. a, b, c, d

Answer 1

The correct option is D a, b, c, d


Potential at point P is greater than at point Q as distance of P is nearer as compared to Q.
And we are taking charge from lower potential to higher potential, but in reality, work done is negative when we take it from higher to lower potential
, so work done will be positive.
Hence work done in taking charge particles from Q to P is positive.
In moving a small negative charge from B to A, work has to be done by the external agency. It is positive.
Due to force of repulsion on the negative charge, velocity decreases and hence the kinetic energy decreases in going from B to A.
Due to the force of attraction of the positive charge, velocity increases and hence the kinetic energy increases in going from Q to P.