Chapter 30, Problem 54A

To determine

To calculate:

Electric force of repulsion between two protons.

Answer to Problem 54A

Electric force of repulsion in between two protons is $5.774\times {10}^{-59}\text{N}$ .

Explanation of Solution

Given:

Distance between the charges $\left(r\right)=2.0\times {10}^{-15}\text{m}$

Coulomb’s law constant $\left(k\right)=9\times {10}^9{\text{Nm}}^{\text{2}}{\text{/C}}^{\text{2}}$

Charge on proton $\left(q_1\right)=1.602\times {10}^{-19}\text{C}$

Charge on proton $\left(q_2\right)=1.602\times {10}^{-19}\text{C}$

Formula Used:

Electric force is given by formula,

  $F_E=k\left(\frac{q_1\times q_2}{r^2}\right)$

Where,

  $F_E:$ Electric force

  $k:$ Coulomb’s law constant

  $q_1:$ Charge on proton

  $q_2:$ Charge on proton

  $r:$ Distance between the charges

Calculation:

Electric force of repulsion between two protons can be calculated by formula,

  $F_E=k\left(\frac{q_1\times q_2}{r^2}\right)$

Substituting the values in above formula,

  $\begin{array}{l}{F}_E=\left(9\times {10}^9{\text{Nm}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\times \left(\frac{\left(1.602\times {10}^{-19}\text{C}\right)\times \left(1.602\times {10}^{-19}\text{C}\right)}{{\left(2.0\times {10}^{-15}\text{m}\right)}^2}\right)\\ =\left(9\times {10}^9{\text{Nm}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\times \left(\frac{{\left(1.602\times {10}^{-19}\text{C}\right)}^2}{{\left(2.0\times {10}^{-15}\text{m}\right)}^2}\right)\\ =\left(9\times {10}^9{\text{Nm}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\times \left(\frac{2.5664\times {10}^{-38}\text{C}}{4\times {10}^{-30}\text{m}}\right)\\ =\left(9\times {10}^9{\text{Nm}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\times \left(6.416\times {10}^{-69}\right)\\ =5.774\times {10}^{-59}\text{N}\end{array}$

Conclusion:

Hence,electric force of repulsion between two protons is $5.774\times {10}^{-59}\text{N}$ .