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Standard XII

Physics

Electric Field Due to a Sphere and Thin Shell

An early mode...

Question

An early model for an atom considered it to have a positively charged nucleus of charge $Z e$ surrounded 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, where $r > R$ ?

\frac{1}{4 π ϵ_{0}} \frac{Z e}{r^{2}}

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\frac{1}{4 π ϵ_{0}} \frac{Z e}{R^{2}}

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\frac{1}{4 π ϵ_{0}} \frac{Z e}{r^{2} - R^{2}}

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Zero

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Solution

The correct option is C Zero
We know that $E$ due to a uniform spherically symmetric charge distribution is the same as $E$ due to the total charge placed at centre of the distribution,
Therefore,
$E = \frac{k Q}{r^{2}} - \frac{k Q}{r^{2}} = 0$ for r > R, where k = $\frac{1}{4 π ϵ_{0}}$ .

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Similar questions

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

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(r < R)

Q. An atom is modelled as a point charge of +e at the nucleus with the electron charge distribution, described by the charge density

ρ = \frac{- 15 e}{8 π a^{3}} (1 - \frac{r^{2}}{a^{2}})

for

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ρ ≃ 0

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where 'a' is a constant.

The net charge contained within a sphere of radius r < a is :

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and volume charge density

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is enclosed by a hollow sphere of radius

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is a positive constant and r is the distance from the center of the sphere.Then the ratio

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and volume charge density

p = \frac{p_{0}}{r}

is enclosed by a hollow sphere of radius

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with negative surface charge density

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and volume charge density

ρ = \frac{ρ_{0}}{r}

is enclosed by a hollow sphere of radius

R_{2}

with negative surface charge density

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An early model for an atom considered it to have a positively charged nucleus of charge Ze surrounded 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, where r>R?

The correct option is C ZeroWe know that E due to a uniform spherically symmetric charge distribution is the same as E due to the total charge placed at centre of the distribution,Therefore,E=kQr2−kQr2=0 for r > R, where k = 14πϵ0 .

An early model for an atom considered it to have a positively charged nucleus of charge $Z e$ surrounded 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, where $r > R$ ?

The correct option is C Zero
We know that $E$ due to a uniform spherically symmetric charge distribution is the same as $E$ due to the total charge placed at centre of the distribution,
Therefore,
$E = \frac{k Q}{r^{2}} - \frac{k Q}{r^{2}} = 0$ for r > R, where k = $\frac{1}{4 π ϵ_{0}}$ .