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Suppose the charge of a proton and an electron differ slightly. One of them is –e, the other is $(e + \Delta e)$. 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 $\Delta e$ is of the order of [Given mass of hydrogen $m_h = 1.67 \times 10^{-27}$ kg]
Explanation
Each H atom carries net charge Δe; equate the Coulomb and gravitational forces.
Detailed Solution
$F_e = F_g$
$\frac{1}{4\pi\varepsilon_0}\frac{(\Delta e)^2}{d^2} = \frac{Gm_h^2}{d^2}$
$9\times10^{9}(\Delta e)^2 = 6.67\times10^{-11}\times1.67\times10^{-27}\times1.67\times10^{-27}$
$(\Delta e)^2 = \frac{6.67\times1.67\times1.67}{9}\times10^{-74}$
$\Delta e \approx 10^{-37}$ C
