The distance between a proton and electron both having a charge $1.6 \times {10^{ - 19}}\,coulomb$, of a hydrogen atom is ${10^{ - 10}}\,metre$. The value of intensity of electric field produced on electron due to proton will be
The distance between a proton and electron both having a charge $1.6 \times {10^{ - 19}}\,coulomb$, of a hydrogen atom is ${10^{ - 10}}\,metre$. The value of intensity of electric field produced on electron due to proton will be
- A
$2.304 \times {10^{ - 10}}\,N/C$
- B
$14.4\,V/m$
- C
$16\,V/m$
- D
$1.44 \times {10^{11}}\,N/C$
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- [JEE MAIN 2024]
What is the magnitude of a point charge which produces an electric field of $2\, N/coulomb$ at a distance of $60\, cm$ $(1/4\pi {\varepsilon _0} = 9 \times {10^9}\,N - {m^2}/{C^2})$
What is the magnitude of a point charge which produces an electric field of $2\, N/coulomb$ at a distance of $60\, cm$ $(1/4\pi {\varepsilon _0} = 9 \times {10^9}\,N - {m^2}/{C^2})$