Chapter 21, Problem 66A

a.

To determine

Force on an electron.

Answer to Problem 66A

  $F=-1.6\times {10}^{-14}\text{N}$ .

Explanation of Solution

Given:

Electrons accelerated in electric field.

Strength of electric field $E=1\times {10}^5\text{N/C}$ directed towards the south.

Formula Used:

As strength of electric field is given by,

  $E=\frac{F}{q}$

Where, $E$ is electric field strength, $F$ is force experienced by charge in electric field and $q$ is the test charge.

Calculation:

As, charge on electron is $-1.602\times {10}^{-19}\text{C}$

So, on putting the given values in above formula,

  $\begin{array}{c}1\times {10}^5=\frac{F}{\left(-1.602\times {10}^{-19}\right)}\\ F=\left(1\times {10}^5\right)\times \left(-1.602\times {10}^{-19}\right)\text{N}\\ =-1.6\times {10}^{-14}\text{N}\end{array}$

Conclusion:

Therefore, the force acting on the electron is $F=-1.6\times {10}^{-14}\text{N}$ .

b.

To determine

Acceleration of the electron.

Answer to Problem 66A

  $a=-1.76\times {10}^{16}{\text{m/s}}^2$

Explanation of Solution

Given:

Field is constant.

Mass of electron is $m=9.11\times {10}^{-31}\text{kg}$

Formula used:

The force is given as:

  $F=ma$

Where, $F$ is force , $m$ is mass and $a$ is acceleration.

Calculation:

As, $F=-1.6\times {10}^{-44}\text{N}$ from previous subpart.

So, on substituting the values

  $\begin{array}{c}-1.6\times {10}^{-14}=9.11\times {10}^{-31}\times a\\ a=\frac{-1.6\times {10}^{-14}}{9.11\times {10}^{-31}}\\ =-1.76\times {10}^{16}{\text{m/s}}^2\end{array}$

Conclusion:

Therefore, the acceleration of the electron is $a=-1.76\times {10}^{16}{\text{m/s}}^2$