First slide

Coulumbs law

Question

Suppose the charge of a proton and an electron differ slightly. One of them is -e, the other is $(e + Δ e)$ . 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 $Δ e$ is of the order of Given mass of hydrogen $[m_{h} = 1.67 \times 10^{- 27} kg]$

Difficult

A

$10^{- 23} C$
B

$10^{- 37} C$
C

$10^{- 47} C$
D

$10^{- 20} C$

Solution

A hydrogen atom consists of an electron and a proton.
$∴ Charge on one hydrogen atom$
$= q_{e} + q_{p} = - e + (e + Δ e) = Δ e$
$Since a hydrogen atom carry a net charge Δ e,$
$∴ Electrostatic force, F_{e} = \frac{1}{4 π ε_{o}} \frac{(Δ e)^{2}}{d^{2}} . . . . (i)$
will act between two hydrogen atoms.
The gravitational force between two hydrogen atoms is given as
$F_{g} = \frac{G m_{h} m_{h}}{d^{2}} . . . . (i i)$
Since, the net force on the system is zero, $F_{e} = F_{g}$ Using eqns.
(i) and (ii), we get
$\frac{(Δ e)^{2}}{4 π ε_{o} d^{2}} = \frac{G m_{h}^{2}}{d^{2}}$
$(Δ e)^{2} = 4 π ε_{o} G m_{h}^{2} = 6.67 \times 10^{- 11} \times {(1.67 \times 10^{- 27})}^{2} / (9 \times 10^{9})$
$Δ e \approx 10^{- 37} C$

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