Question

Figure shows tracks of three charged particles in a uniform electrostatic field. Which particle has the highest magnitude of charge to mass ratio (Assume gravity to be negligible) ?

• A. $1$
• B. $2$
• C. $3$
• D. All particles have same charge to mass ratio

Answer 1

The correct option is C $3$

Since gravity is negligible here, so the particles move under the influence electrostatic field $\left(E\right)$.

Electrostatic force $\left(F\right)$ acting on the particle can be written as,

$F=qE$

From the Newton's Second Law of Motion, we can write

$ma=qE$

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

Here, deviation is nothing but the displacement produced in the $y-$direction due to the electrostatic force on the charged particle in the given electric field.

deviation$=\Delta d=u_{y}t+\frac{1}{2}a{t}^2$

Substituting the value of $a$,

$\Rightarrow \Delta d=u_{y}t+\frac{1}{2}\left(\frac{qE}{m}\right)t^2$

We can see that $u_y=0$ for all the three charged particles, as there is no initial velocity in the $\text{y-direction}$ when the particles enter the electric field.

$\Rightarrow \Delta d=\frac{qE}{2m}{t}^2$

$\Rightarrow \Delta d\propto \frac{q}{m}$

Thus, greater the magnitude of (charge/mass) ratio, greater will be the deviation.

Hence, particle $\left(3\right)$ has the highest magnitude of charge to mass ratio as it has the largest deflection.

Note: As the particle $1$ and $2$ are deviated towards the positive plate, they are negatively charged, while particle 3 is positively charged.