Question

Figure shows three arrangements of electric field lines. In each arrangement, a proton is released from rest at point $ A $ and then accelerated through point $ B $ by the electric field. Points $ A $ and $ B $ have equal separations in the three arrangements. If $ p_{1}, p_{2} $ and $ p_{3} $ are linear momentum of the proton at point $ B $ in the three arrangement respectively, then

• A. $ p_{1} > \, p_{3} > \, p_{2} $
• B. $ p_{1} > \, p_{2} > \, p_{3} $
• C. $ p_{2} > \, p_{1} > \, p_{3} $
• D. $ p_{3} > \, p_{2} > \, p_{1} $

Answer 1

The Correct Option is B

The region in which electric lines of forces are closer, have more value of electric field than the region in which electric lines of forces are farther
(i) Electric lines of forces are closer and uniform, so acceleration on proton is increasing till $B$, hence proton has maximum velocity at $B$, therefore $p_{1}$ is maximum
(ii) In this figure, electric lines of forces are going away from each other, hence electric field continuously decreases, so $p_{2}$ is moderate at $B$
(iii) In this figure, electric lines of forces are farthest among till three figures, hence acceleration on proton is minimum due to weakest electric field, hence $p_{3}$ is minimum at $B$
$\therefore p_{1}>\,p_{2}>\,p_{3}$