Question

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.

Short answer

The rank of the protons according to the magnitude of accelerations due to the field is $c>e>b>d>a$.

Step-by-step solution

The given data

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

Understanding the concept of acceleration

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)

Calculation of the rank of protons according to their acceleration

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 $c>e>b>d>a$.