Question

Electric field given by the vector
_ _ _
E = x i + y j is
present in the XY plane. A small ring carrying charge
+Q, which can freely slide on a smooth non conducting
rod, is projected along the rod from the point (0, L)
such that it can reach the other end of the rod. What
minimum velocity should be given to the ring ? (Assume
zero gravity)

Answer 1

Concept :- For minimum velocity . the position where we get electric field intensity zero , body have require to reach atleast that point for reaching other end of the rod. Means we have to find potential difference between intial point to the point where electric field intensity will be zero.

Arrangement is shown in figure , in case of rod , electric field intensity will be zero at midpoint . So, midpoint of rod is (L/2,L/2) where electric field will be zero.
We know, ∆V = -∫E.dr
Let the position vector of object , dr = dx i + dy j [we should not use direction of position vector , because question didn't consider about direction ]
Given, E = x i + y j
Now, ∆V = - ∫(x i + y j ).(dx i + dy j)
∆V = -∫xdx - ∫ydy
∆V = $-\int\limits^{L/2}_0{x}\,dx+-\int\limits^{L/2}_{L}{y}\,dy$
= $-1/2[x^2]^{L/2}_0-1/2[y^2]^{L/2}_L$
= - L²/8 + 3L²/8
= 2L²/8 = L²/4

Now, work done = kinetic energy
Q∆V = 1/2mv²
⇒QL²/4 = 1/2 mv²
⇒ QL²/2m = v²
Hence, v = √{QL²/2m}