Rankers Physics

Coulomb's Law: Practice Problem & Solution

2. 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 $m_h = 1.67 \times 10^{-27} \text{ kg}$] (2017-Delhi)
$10^{-23} \text{ C}$
$10^{-37} \text{ C}$
$10^{-47} \text{ C}$
$10^{-20} \text{ C}$

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Coulomb's Law. Understanding the underlying formula is key to arriving at the correct answer below:

Net charge on each atom is $\Delta e$. Equating electrostatic repulsion to gravitational attraction: $\frac{1}{4\pi\epsilon_0} \frac{(\Delta e)^2}{d^2} = \frac{G m_h^2}{d^2}$. Solving yields $\Delta e \approx 10^{-37} \text{ C}$.

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