The distance between a proton and an electron | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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:

Similar Questions

The distance between two charges $25\,\mu C$ and $36\,\mu C$ is $11\,cm$. At what point on the line joining the two charges will the electric field intensity be zero?

As shown in the figure,at what distance in $cm$ from point $A$ will the electric field be zero?

Two equal negative charges $-q$ are fixed at points $(0, -a)$ and $(0, a)$ in the $x-y$ plane. $A$ positive charge $Q$ is released from rest at a point $(2a, 0)$. The charge $Q$ will

Two charged conducting spheres of radii $5 \ cm$ and $10 \ cm$ have equal surface charge densities. If the electric field on the surface of the smaller sphere is $E$,then the electric field on the surface of the larger sphere is

Vedclass Test Series

Exam Paper Generator

Online Exam Module