Understand the setup

• A charged particle of mass $m$ and charge $q$ is attached to a massless string of equilibrium length $l$.
• One end is fixed at $O$, the particle starts at point $A$.
• The motion is on a frictionless horizontal plane.
• A uniform electric field is switched on.
• We need the speed of the particle when it crosses the $x$-axis.

Key idea: use work-energy theorem

The only force doing work is the electric force.

• Tension in the string is always along the string and does no work in the constrained motion.
• Hence, $\Delta K = W_{\text{electric}}$

Initial and final positions

From the standard figure for this problem, the particle initially is at point $A$ on the positive $y$-axis, with $OA = l$ so initially its coordinates are $A=(0,l).$

When it crosses the $x$-axis while remaining at distance $l$ from $O$, it is at the point $B=(l,0).$

Thus the displacement in the direction of electric field (along $+x$) is $\Delta x = l - 0 = l.$

Work done by electric field

Electric force on the charge is $F = qE$ along the field direction.

Therefore work done from $A$ to $B$ is $W = qE\,\Delta x = qEl.$

Apply energy conservation

Initially the system is at rest, so $K_i = 0.$

At the instant it crosses the $x$-axis, let speed be $v$. Then $K_f = \frac12 mv^2.$

Hence, $\frac12 mv^2 = qEl.$

Solving, $v^2 = \frac{2qEl}{m}$ $v = \sqrt{\frac{2qEl}{m}}.$

Match with the options

This corresponds to: $\boxed{\sqrt{\frac{2qEl}{m}}}$

So the correct option is D.

Comparison with stored answer

Stored correct answer is C: $\sqrt{\frac{qEl}{m}}$.

But from work-energy theorem, the correct result is clearly $\boxed{\sqrt{\frac{2qEl}{m}}}.$

Therefore, I disagree with the stored answer.