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(D) The electric field $E$ produced by a point charge $q$ at a distance $r$ is given by the formula: $E = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q}{

Vedclass · Accepted Answer

$1.44 	imes {10^{11}} \ N/C$

Vedclass · Answer

(D) The electric field $E$ produced by a point charge $q$ at a distance $r$ is given by the formula: $E = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q}{r^2}$ Given: Charge $q = 1.6 	imes {10^{ - 19}} \ C$ Distance $r = {10^{ - 10}} \ m$ Coulomb's constant $k = \frac{1}{4\pi \varepsilon_0} = 9 	imes {10^9} \ N \cdot m^2/C^2$ Substituting the values: $E = (9 	imes {10^9}) 	imes \frac{1.6 	imes {10^{ - 19}}}{({10^{ - 10}})^2}$ $E = 9 	imes 1.6 	imes {10^9} 	imes {10^{ - 19}} 	imes {10^{20}}$ $E = 14.4 	imes {10^{10}} \ N/C$ $E = 1.44 	imes {10^{11}} \ N/C$ Thus,the correct option is $D$.

The distance between a proton and an electron in a hydrogen atom is ${10^{ - 10}} \ m$. Both have a charge of magnitude $1.6 \times {10^{ - 19}} \ C$. The magnitude of the electric field intensity produced at the position of the electron due to the proton is:

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