Chapter 16, Problem 61GP

(a)

To determine

The distance covered by the electron before it stops.

Answer to Problem 61GP

  $\text{11}{\text{.55×10}}^{\text{-2}}\text{m}$

Explanation of Solution

Given:

The speed of the electron is $21.5\times {10}^6\text{m/s}$ .

The magnitude of electric field is $11.4\times {10}^3\text{N/C}$ .

The electron is travelling parallel to the electric field.

Formula used:

  $\text{F = m}\cdot \text{a}$

Where,

  $\text{F}$ is the force on the electron,

  $\text{m}$ is the mass of electron which is $\text{9}\text{.1}\times {\text{10}}^{-31}\text{kg}$

  $\text{a}$ is the acceleration.

  $\text{F = q}\cdot \text{E}$

Where,

  $\text{q}$ is the charge on an electron which is $\text{1}{\text{.6×10}}^{\text{-19}}\text{C}$ .

  $\text{E}$ is the magnitude of the electric field applied.

  ${\text{v}}^{\text{2}}{\text{= u}}^{\text{2}}\text{- 2aS}$

Where,

  $\text{v}$ is the final velocity,

  $\text{u}$ is the initial velocity and

  $\text{S}$ is the displacement of the electron.

Calculation:

The acceleration can be calculated by equating the force equations.

  $\begin{array}{l}\text{F}\text{=}\text{qE}\\ \text{ma}\text{=}\text{qE}\end{array}$

  $\text{a =}\frac{\text{qE}}{\text{m}}$

  $\text{a}\text{=}\frac{\left(\text{1}{\text{.6×10}}^{\text{-19}}\right)\text{×}\left(\text{11}{\text{.4×10}}^{\text{3}}\right)}{\left(\text{9}{\text{.1×10}}^{\text{-31}}\right)}$

  $\text{a}\text{=}{\text{2×10}}^{\text{15}}{\text{m/s}}^{\text{2}}$

Here, the final velocity of the electron is zero. So, the distance travelled by the electron before it stops is

  ${\text{v}}^{\text{2}}{\text{=u}}^{\text{2}}\text{- 2aS}$

  ${\text{0=u}}^{\text{2}}\text{-2aS}$

  $\text{S=}\frac{{\text{u}}^{\text{2}}}{\text{2a}}$

  $\text{S=}\frac{{\left(\text{21}{\text{.5×10}}^{\text{6}}\right)}^{\text{2}}}{\text{2}\left({\text{2×10}}^{\text{15}}\right)}$

  $\text{S=11}{\text{.55×10}}^{\text{-2}}\text{m}$

(b)

To determine

The time before the electron returns to its starting point.

Answer to Problem 61GP

  $\text{2}{\text{.15×10}}^{\text{-8}}\text{s}$

Explanation of Solution

Given:

The speed of the electron is $21.5\times {10}^6\text{m/s}$ .

The magnitude of electric field is $11.4\times {10}^3\text{N/C}$ .

The acceleration is ${\text{2×10}}^{\text{15}}{\text{m/s}}^{\text{2}}$ .

The electron is travelling parallel to the electric field.

Formula used:

  $\text{v = u + at}$

Where,

  $\text{v}$ is the final velocity of the electron.

  $\text{u}$ is the initial velocity of the electron.

  $\text{a}$ is the acceleration on the electron.

  $\text{t}$ is the time period.

Calculation:

The time that the electron will take to reach its initial position is twice the time it take to reach zero velocity.

  $\text{v = u + at}$

  $\text{0 = u +}\left(\text{-at}\right)$

  $\text{u = at}$

  $\text{t =}\frac{\text{u}}{\text{a}}$

  $\text{t =}\frac{\text{21}{\text{.5×10}}^{\text{6}}}{{\text{2×10}}^{\text{15}}}$

  $\text{t =1}{\text{.075×10}}^{\text{-8}}\text{s}$

The time taken to reach the initial value is

  $\begin{array}{l}\text{2t = 2×}\left(\text{1}{\text{.075×10}}^{\text{-8}}\right)\text{s}\\ \text{2t = 2}{\text{.15×10}}^{\text{-8}}\text{s}\end{array}$