Question 873: Electrostats & Capacitance

Question

In hydrogen atom an electron revolves around a proton (in nucleus) at a distance 'r' m. the intensity of electric field due to the proton at distance 'r' is $$5 	imes 10^{11} \mathrm{NC}^{-1}$$, the magnitude of force between the electron and proton is [charge on electron $$=1.6 	imes 10^{-19} \mathrm{C}$$]

Accepted Answer

To find the magnitude of the force between the electron and proton, we can use the relationship between the electric field and the force on a charge. The force $$\mathbf{F}$$ on a charge $$\mathbf{q}$$ in an electric field $$\mathbf{E}$$ is given by:
$$ \mathbf{F} = q \mathbf{E} $$
Here, the electric field intensity due to the proton at distance $$r$$ is given as:
$$ E = 5 	imes 10^{11} \mathrm{NC}^{-1} $$
The charge on the electron is:
$$ q = 1.6 	imes 10^{-19} \mathrm{C} $$
Substituting these values into the equation for the force:
$$ \mathbf{F} = (1.6 	imes 10^{-19} \mathrm{C}) 	imes (5 	imes 10^{11} \mathrm{NC}^{-1}) $$
$$ \mathbf{F} = 8 	imes 10^{-8} \mathrm{N} $$
Therefore, the magnitude of the force between the electron and proton is:
Option D:
$$8 	imes 10^{-8} \mathrm{N}$$

Answer

$$4 	imes 10^8 \mathrm{~N}$$

Answer

$$8 	imes 10^8 \mathrm{~N}$$

Answer

$$4 	imes 10^{-8} \mathrm{~N}$$

Answer

$$8 	imes 10^{-8} \mathrm{~N}$$

Electrostats & Capacitance

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