1. Force on the charged particle

A charge $q$ placed in a uniform electric field $E$ experiences a constant force

$F = qE$

So the acceleration is constant:

$a = \frac{F}{m} = \frac{qE}{m}$

(If $q<0$, the direction changes, but the speed-distance relation still depends on magnitude appropriately.)

2. Use kinematics / work-energy relation

Since the particle is released from rest,

$u=0$

For constant acceleration, we use

$v^2 = u^2 + 2ax$

Thus,

$v^2 = 2ax = 2\left(\frac{qE}{m}\right)x$

So,

$v = \sqrt{\frac{2qE}{m}\,x}$

In magnitude form,

$v \propto \sqrt{x}$

3. Nature of the graph

The relation $v \propto \sqrt{x}$ means:

• $v=0$ at $x=0$
• $v$ increases with $x$
• slope decreases as $x$ increases

So the graph is an increasing curve, concave downward.

4. Choose the correct option

Hence the correct schematic graph is the one representing

$v \text{ vs } x : \quad v \propto \sqrt{x}$

This corresponds to Option B.

5. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They match.