Chapter 16, Problem 54GP

To determine

To Find: Amount of electric field $\left(\overrightarrow{\mathrm{E}}\right)$ to suspend a proton at rest in electric field under action of gravity

Answer to Problem 54GP

  $1.02\times {10}^{-7}\text{N/C,upwards}$

Explanation of Solution

Given:

Mass of proton: $1.67\times {10}^{-27}\text{kg}$

Electron is suspended in $\left(\overrightarrow{\mathrm{E}}\right)$ against the action of gravity.

Formula used:

  $\mathrm{E}=\frac{\mathrm{m}\mathrm{g}}{\mathrm{q}}$

  $\text{Where,}\mathrm{m}=$ mass of proton

  $\mathrm{g}=$ acceleration due to gravity

  $\mathrm{q}=$ charge on electron

Calculation:

Mass of Proton $(\mathrm{m})=1.67\times {10}^{-27}\text{kg}$

  $\mathrm{g}=9.8{\text{m/s}}^{\text{2}}$

Charge on proton $(\mathrm{e})=1.6\times {10}^{-19}\text{C}$

The electric field $\mathrm{E}=\frac{\mathrm{F}}{\mathrm{e}}=\frac{\mathrm{m}\mathrm{g}}{\mathrm{e}}$

  $\begin{array}{l}=\frac{\left(1.67\times {10}^{-27}\right)(9.8)}{\left(1.6\times {10}^{-19}\right)}\text{N/C}\hfill \\ \mathrm{E}=1.02\times {10}^{-7}\text{N}/\text{C}\hfill \end{array}$

Conclusion:

Amount of electric field required to suspend a proton at rest against action of gravity is $1.02\times {10}^{-7}\text{N/C}$ (in upward direction)