Figure 22-33 shows five protons that are launched in a uniform electric field $\overrightarrow{E}$; the magnitude and direction of the launch velocities are indicated. Rank the protons according to the magnitude of their accelerations due to the field, greatest first.

Figure 22-33 showing five protons with their magnitude and direction of launch in uniform electric field are given.

The acceleration of a body is determined by the change in the velocity during a given time. Thus, using the direct relation of acceleration to velocity, we can get the acceleration of the given cases.

Formula:

The acceleration of a charge particle inside a uniform electric field,

$a=(v\cos \theta +v\sin \theta )t$(i)

For proton a:

The acceleration value is given using equation (i) as:

$\begin{array}{c}{a}_1=(10\cos {135}^0+10\sin {135}^0)t\\ =-10\sqrt{2}t\\ =-14.14t{\text{m/s}}^2\end{array}$

For proton b:

The acceleration value is given using equation (i) as:

$\begin{array}{c}{a}_2=(3\cos 0^0+3\sin 0^0)t\\ =3t{\text{m/s}}^2\end{array}$

For proton c:

The acceleration value is given using equation (i) as:

$\begin{array}{c}{a}_3=(16\cos {90}^0+16\sin {90}^0)t\\ =16t{\text{m/s}}^2\end{array}$

For proton d:

The acceleration value is given using equation (i) as:

$\begin{array}{c}{a}_4=(5\cos {180}^0+5\sin {180}^0)t\\ =-5t{\text{m/s}}^2\end{array}$

For proton e:

The acceleration value is given using equation (i) as:

$\begin{array}{c}{a}_5=(7\cos {45}^0+7\sin {45}^0)t\\ =7\sqrt{2}t\\ =9.89t{\text{m/s}}^2\end{array}$

Hence, the rank of the protons according to their acceleration is $c>e>b>d>a$.