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 × 1 0 11 NC − 1

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 $5 \times 1 0^{11} NC^{- 1}$ , the magnitude of force between the electron and proton is [charge on electron $= 1.6 \times 1 0^{- 19} C$ ]

Options:

Answer:

D

Solution:

$F = qE = 1.6 \times 1 0^{- 19} \times 5 \times 1 0^{11} = 8 \times 1 0^{- 8} N$

TestHub · Accepted Answer

​ F = qE = 1.6 × 1 0 − 19 × 5 × 1 0 11 = 8 × 1 0 − 8 N ​

Physics - Electrostatics Question with Solution | TestHub

Options:

Doubts & Discussion