Q2P - Fundamentals Of Physics

Question

In Fig. 22-34 the electric field lines on the left have twice the separation of those on the right. (a) If the magnitude of the field at A is$\mathbf{40}\mathbf{ }\mathit{N}\mathbf{/}\mathit{C}$, what is the magnitude of the force on a proton at A? (b) What is the magnitude of the field at B?

Step-by-Step Solution

Answer

1. The magnitude of the force on a proton is $6.4x{10}^{-18} \mathrm{N}$
2. The magnitude of the field at B is $20 \mathrm{N}/\mathrm{C}$

Step 1: The given data

1. The electric charge of the proton, ${\mathrm{q}}_{\mathrm{p}}=1.6\times {10}^{-19} \mathrm{C}$
2.  Strength of the electric field at A with direction,  ${\mathrm{E}}_{\mathrm{A}}=(-40 \mathrm{N}/\mathrm{C})\hat{\mathrm{i}}$
3.  The electric field lines on the left are twice to that of the right. 

Step 2: Understanding the concept of electric field

Using the concept of electric field relation with force per unit charge, we can get the required value of the force and the electric field at B.

 

Formula: 

 

The force on a charged particle in an electric field, $\overrightarrow{F}=q_{0 }\mathrm{E}$                                (1)

where ${\mathrm{q}}_0$= electric charge on the particle, and $\overrightarrow{\mathrm{E}}$= Strength of the electric field

Step 3: a) Calculation of the magnitude of the force at A

As the force at A is orienting our x-axis rightward (so$\hat{\mathrm{i}}$points right in the figure), we can calculate the electric field at A using equation (1) as follows:

$$\begin{gathered} \overrightarrow{F}=(+1.6\times {10}^{-19 }\mathrm{C})(-40 \mathrm{N}/\mathrm{C} \hat{\mathrm{i}}) \\ =(-6.4\times {10}^{-18} \mathrm{N})\hat{\mathrm{i}} \end{gathered}$$

Hence, the magnitude of the force on the proton is $64x{10}^{-18} \mathrm{N}$and its direction $\left(-\stackrel{⏜}{\mathrm{i}}\right)$ is leftward.

Step 4: b) Calculation of the magnitude of the electric field at B

The field strength is proportional to the “crowdedness” of the field lines. So, from the given data, we conclude that 

$\begin{array}{l}{\mathrm{E}}_{\mathrm{A}}=2{\mathrm{E}}_{\mathrm{B}}\\ {\mathrm{E}}_{\mathrm{B}}=\frac{{\mathrm{E}}_{\mathrm{A}}}{2}\\ =20 \mathrm{N}/\mathrm{C}\end{array}$

Hence, the magnitude of the electric field is $20 \mathrm{N}/\mathrm{C}$