Name: Electric and Gravitational Forces
Uploaded: 2019-07-19
Duration: 3 min 11 s
Description: Do you know the difference and similarities between the gravitational force and electrostatic force? Watch this video to explore it. Don't miss it.

Question

An early model of an atom considered it to have a positively charged point nucleus of charge

Z e

, surrounded

(r > R)

by a uniform density of negative charge up to a radius

R

. The atom as a whole is neutral. For this model, what is the electric field at a distance

r

from the nucleus?

Accepted Answer

Figure shows the charge distribution for the given model of the atom.

As the atom is neutral, the total negative charge in a sphere of radius $R$ must be $- Z e$ . If $p$ is the negative charge density, then we must have

$Z e + \frac{4 π R^{3}}{3} ρ = 0$ or $ρ = - \frac{3 Z e}{4 π R^{3}}$

By spherical symmetry of the charge distribution, the electric field $\vec{E}$ depends only on radial distance $r$ and not on the direction of $\vec{r}$ . It should point radially inwards or outwards. So we imagine a spherical Gaussian surface of radius $r$ centered at the nucleus.

(i) For $r < R$ , Flux through the Gaussian surface,

$ϕ_{E} = E \times 4 π r^{2}$

Charge enclosed by the Gaussian surface,

$q =$ Positive nuclear charge $+$ Negative charge in a sphere of radius $r$

$= Z e + \frac{4 π r^{3}}{3} ρ = Z e + \frac{4 π r^{3}}{3} \cdot (- \frac{3 Z e}{4 π R^{3}})$

$= Z e (1 - \frac{r^{3}}{R^{3}})$

Applying Gauss's theorem, $ϕ_{E} = q / ε_{0}$ , we get

$E \times 4 π r^{2} = \frac{Z e}{ε_{0}} (1 - \frac{r^{3}}{R^{3}})$

or

$E = \frac{Z e}{4 π ε_{0}} (\frac{1}{r^{2}} - \frac{r}{R^{3}})$ $(r < R)$

The field $E$ points radially outward.

(ii) For $r > R$ . As the atom is neutral, the total charge enclosed by the Gaussian surface is zero. By Gauss's theorem,

$E \times 4 π r^{2} = 0$

$E = 0$ . $(r > R)$

(iii) At $r = R$ . Both of the above cases give the same result:

$E = 0$ .

An early model of an atom considered it to have a positively charged point nucleus of charge $Z e$ , surrounded $(r > R)$ by a uniform density of negative charge up to a radius $R$ . The atom as a whole is neutral. For this model, what is the electric field at a distance $r$ from the nucleus?

Important Questions on Electric Charges and Field

Learn and Prepare for any exam you want !

An early model of an atom considered it to have a positively charged point nucleus of charge Ze, surrounded (r>R) by a uniform density of negative charge up to a radius R. The atom as a whole is neutral. For this model, what is the electric field at a distance r from the nucleus?

Important Questions on Electric Charges and Field

Learn and Prepare for any exam you want !

An early model of an atom considered it to have a positively charged point nucleus of charge $Z e$ , surrounded $(r > R)$ by a uniform density of negative charge up to a radius $R$ . The atom as a whole is neutral. For this model, what is the electric field at a distance $r$ from the nucleus?