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Electric Field Due to a Sphere and Thin Shell

In one model ...

Question

In one model of the electron, the electron of mass $m_{e}$ is thought to be a uniformly charged shell of radius $R$ and total charge $e$ , whose electrostatic energy $E$ is equivalent to its mass $m_{e}$ via Einstein's mass energy relation $E = m_{e} c^{2}$ . In this model, $R$ is approximately ( $m_{e} = 9.1 \times 10^{- 31} k g$ , $c = 3 \times 10^{8} m s^{- 1}$ , $1 / 4 π ε_{0} = 9 \times 10^{9} F a r a d s m^{- 1}$ , magnitude of the electron charge $= 1.6 \times 10^{- 19} C$ )

1.4 \times 10^{- 15} m

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2 \times 10^{- 13} m

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5.3 \times 10^{- 11} m

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2.8 \times 10^{- 35} m

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Solution

The correct option is A $1.4 \times 10^{- 15} m$
Comparing the electrostatic energy of a shell with radius R with $m c^{2}$
So we get $\frac{e^{2}}{8 π ϵ R} = m c^{2}$
on putting value we get $R = 1.4 \times 10^{- 15} m$

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Similar questions

5.3 \times 10^{- 11} m

= 9.1 \times 10^{- 31} k g, m p = 1.67 \times 10^{- 27} k g, e = 1.6 \times 10^{- 19} C

G = 6.67 \times 10^{- 11} N m^{2} k g^{- 2}

3 \times 10^{10} m s^{- 1}

0.2 T

e = 1.6 \times 10^{- 19} C

m_{e} = 9.1 \times 10^{- 31} k g

0.55 T

m_{e} = 9.1 \times 10^{- 31} k g

e = 1.6 \times 10^{- 19} C

- 1.6 \times 10^{- 19} C

e / m

1.758820 \times 10^{11} C/kg

e

m_{e}

9.1 \times 10

^{- 31} k g

e = 1.6 \times 10

^{- 19} C

G = 6.6 \times 10

^{- 11}

N m^{2} / k g^{2}

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