Chapter 16, Problem 42P

(a)

To determine

The direction required for the electric field.

Explanation of Solution

Introduction:

The electron will experience a force in the opposite direction of the electric field.

Since, the electron is to be brought to rest. The electric field must be in the same direction as the initial velocity of the electron. So, the required direction is right.

Conclusion:

Thus, the required direction is right.

(b)

To determine

The strength of the electric field.

Answer to Problem 42P

The strength of the electric field is $640.5\text{N/C}$ .

Explanation of Solution

Given info:

The mass of the electron is $m=9.11\times {10}^{-31}\text{kg}$ .

The charge on electron is $q=1.6\times {10}^{-19}\text{C}$ .

The electric field is $E=1.45\times {10}^4\text{N/C}$ .

The space is $S=4\text{cm}$ .

Formula used:

The initial speed of the electron is $v_0=0.01c$ .

The final speed of the electron is $v=0$ .

The negative acceleration is,

  $a=\frac{eE}{m_e}$ … (1)

So,

  $\begin{array}{c}{v}^2=v_0^2-2as\\ 0=v_0^2-2as\\ a=\frac{v_0^2}{2s}\end{array}$ … (2)

Equate the equation (1) and (2).

  $\begin{array}{c}\frac{eE}{m_e}=\frac{v_0^2}{2s}\\ E=\frac{m_e{\left(0.01c\right)}^2}{2es}\end{array}$

Calculation:

Substituting the given values, we get

  $\begin{array}{c}E=\frac{\left(9.11\times {10}^{-31}\right){\left(0.01\times 3\times {10}^8\right)}^2}{2\left(1.6\times {10}^{-19}\right)\left(4\times {10}^{-3}\right)}\text{N/C}\\ =640.5\text{N/C}\end{array}$

Conclusion:

Thus, the strength is $640.5\text{N/C}$ .