Question

In a hydrogen atom in its ground state, the electron is on average a distance of about $\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{m}$from the proton. What is the magnitude of the electric field due to the proton at this distance from the proton?

Short answer

The electric field due to proton at the given distance is $5.76\times {10}^{11}\mathrm{N}/\mathrm{C}$.

Step-by-step solution

Identification of given data

The given data can be listed below,

• The distance of electron from the proton is$r=0.5\times {10}^{-10}\mathrm{m}$

Concept/Significance of Coulomb’s Law.

An electrostatic law is Coulomb's law. The other law is that opposite charges attract and comparable charges repel, which is factored into Coulomb's law in its vector form.

Determination of the electric field due to proton

Following coulomb law and definition of electric field, electric field is given by,

$E=\frac{q}{4\pi {\epsilon }_0{r}^2}$

Here,q is the charge on the proton whose value is $1.6\times {10}^{-19}C,\frac{1}{4\pi {\epsilon }_0}$, is the coulomb constant whose value is $9\times {10}^{9}N.m^2{C}^2$ and r is the distance of proton.

Substitute all the values in the above,

$$\begin{gathered} E=\frac{\left(9\times {10}^{9}N.m^2/C^2\right)\left(1.6\times {10}^{-19}C\right)}{{\left(0.5\times {10}^{-10}m\right)}^2} \\ =5.76\times {10}^{11}N/C \end{gathered}$$

Thus, the electric field due to proton at the given distance is $5.76\times {10}^{11}N/C$.