Chapter 16, Problem 56GP

(a)

To determine

Acceleration experienced by an electron.

Answer to Problem 56GP

  $2.6\times {10}^{13}\text{m}/{\text{s}}^{\text{2}}\text{,up}$

Explanation of Solution

Given:

Electric field due to earth is $150\mathrm{N}/\mathrm{C}$ in downward direction

Formula used:

Electron in an electric field experiences a force given by

  $\mathrm{F}=\mathrm{e}\mathrm{E}=\mathrm{m}\mathrm{a}$

  $\Rightarrow$ So acceleration is given by

  ${\mathrm{a}}_{\mathrm{e}}=\frac{\mathrm{e}\mathrm{E}}{\mathrm{m}}$

Calculation:

  ${\mathrm{a}}_{\mathrm{e}}=\frac{\mathrm{e}\mathrm{E}}{\mathrm{m}}=\frac{\left(1.6\times {10}^{-19}\mathrm{C}\right)\left(150\mathrm{N}/\mathrm{C}\right)}{\left(9.11\times {10}^{-31}\mathrm{k}\mathrm{g}\right)}$

  ${\mathrm{a}}_{\mathrm{e}}=2.6\times {10}^{13}\mathrm{m}/{\mathrm{s}}^2$ (upward)

Conclusion:

Electron experience an acceleration of ${\mathrm{a}}_{\mathrm{e}}=2.6\times {10}^{13}\mathrm{m}/{\mathrm{s}}^2$ near the surface of earth

(b)

To determine

Acceleration experienced by proton.

Answer to Problem 56GP

  $1.4\times {10}^{10}\mathrm{m}/{\mathrm{s}}^2$ , down

Explanation of Solution

Given:

Electric field experienced by proton is $150\mathrm{N}/\mathrm{C}$

Formula used:

Force experienced by proton in an electric field ( E ) is given by

  ${\mathrm{F}}_{\mathrm{E}}=\mathrm{e}\mathrm{E}=\mathrm{m}\mathrm{a}$

So acceleration in proton is given by

  ${\mathrm{a}}_{\mathrm{p}}=\frac{\mathrm{e}\mathrm{E}}{{\mathrm{m}}_{\mathrm{p}}}$

Calculation:

  ${\mathrm{a}}_{\mathrm{p}}=\frac{\mathrm{e}\mathrm{E}}{{\mathrm{m}}_{\mathrm{p}}}=\frac{\left(1.6\times {10}^{-19}\mathrm{C}\right)\left(150\mathrm{N}/\mathrm{C}\right)}{\left(1.67\times {10}^{-27}\mathrm{k}\mathrm{g}\right)}$

  ${\mathrm{a}}_{\mathrm{p}}=1.4\times {10}^{10}\mathrm{m}/{\mathrm{s}}^2$

Conclusion:

Proton experience an acceleration of $1.4\times {10}^{10}\mathrm{m}/{\mathrm{s}}^2$ near the surface of earth

(c)

To determine

Ratio of acceleration experienced by electron and proton to the acceleration of gravity

Answer to Problem 56GP

For electron: $2.7\times {10}^{12}$

For proton: $1.5\times {10}^9$

Explanation of Solution

Given:

From part (a) acceleration of electron in $\overrightarrow{\mathrm{E}}$ is $2.6\times {10}^{13}\mathrm{m}/{\mathrm{s}}^2$

From part (b) acceleration of proton in $\overrightarrow{\mathrm{E}}$ is $1.4\times {10}^{10}\mathrm{m}/{\mathrm{s}}^2$

Acceleration due to gravity $=9.8\mathrm{m}/{\mathrm{s}}^2$

Formula used:

Required ratio (for electron) $=\frac{{\mathrm{a}}_{\mathrm{e}}}{\mathrm{g}}$

Required ratio (for proton) $=\frac{{\mathrm{a}}_{\mathrm{p}}}{\mathrm{g}}$

Calculation:

For electron: $=\frac{{\mathrm{a}}_{\mathrm{e}}}{\mathrm{g}}=\frac{2.6\times {10}^{13}\mathrm{m}/{\mathrm{s}}^2}{9.8\mathrm{m}/{\mathrm{s}}^2}=2.7\times {10}^{12}$

For proton $=\frac{{\mathrm{a}}_{\mathrm{p}}}{\mathrm{g}}=\frac{1.4\times {10}^{10}\mathrm{m}/{\mathrm{s}}^2}{9.8\mathrm{m}/{\mathrm{s}}^2}=1.5\times {10}^9$

Conclusion:

The ratio of acceleration experienced by electron near earth surface to the acceleration due to gravity is $2.7\times {10}^{12}$ while same ratio in case of proton is $1.5\times {10}^9$