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  • Can someone help me with this, - Electrostatics - NEET-3

Suppose the charge of a proton and an electron differ slightly. One of them is -e, the other is (e + \Deltae). If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then \Deltae is of the order of [Given mass of hydrogen mh = 1.67 x 10-27 kg]

  • Option 1)

    10-20 C

  • Option 2)

    10-23 C

  • Option 3)

    10-37 C

  • Option 4)

    10-47 C

 

Answers (1)

best_answer

As we learnt in 

Coulombic force -

Fpropto Q_{1}Q_{2}=Fpropto frac{Q_{1}Q_{2}}{r^{2}}=F=frac{KQ_{1}Q_{2}}{r^{2}}

- wherein

K - proportionality Constant 

Q1 and Q2 are two Point charge

 

 Charge on one hydrogen atoms

q_{e}+q_{p}=e+\left ( e+se \right )=se

\therefore F_{e} =\frac{1}{4\pi \varepsilon _{0}} \frac{\left ( se \right )^{2}}{d^{2}}

gravitational force between two hydrogen

F_{g} = \frac{Gm_{n}m_{n}}{d^{2}}

2f force on the system is 2ro

\therefore F_{e}=F_{g} \Rightarrow \frac{\left ( se \right )^{2}}{4\pi \varepsilon _{0}d^{2}} =\frac{Gm^{2}n}{d^{2}}

\left ( se \right )^{2} =4\pi \varepsilon _{0}Gm^{2}n

se = 10^{-37}c

 


Option 1)

10-20 C

Incorrect

Option 2)

10-23 C

Incorrect

Option 3)

10-37 C

Correct

Option 4)

10-47 C

Posted by

divya.saini

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