2020 Phase-2 Electric Charges and FieldsCoulomb's law - force and resulting acceleration Medium Question 14 of 42 in Electric Charges and Fields

The acceleration of an electron due to the mutual attraction between the electron and a proton when they are 1.6 Å apart is, ($m_e=9\times10^{-31}$ kg, $e=1.6\times10^{-19}$ C, $\dfrac{1}{4\pi\varepsilon_0}=9\times10^9\ Nm^2C^{-2}$)

A $10^{25}\ m/s^2$
B $10^{24}\ m/s^2$
C $10^{23}\ m/s^2$
D $10^{22}\ m/s^2$

The answer is D.

Solution

$F=\dfrac{1}{4\pi\varepsilon_0}\dfrac{e^2}{r^2}=9\times10^9\times\dfrac{(1.6\times10^{-19})^2}{(1.6\times10^{-10})^2}=9\times10^9\times(1)^2\times10^{-18}=9\times10^{-9}$ N. $a=\dfrac{F}{m_e}=\dfrac{9\times10^{-9}}{9\times10^{-31}}=10^{22}\ m/s^2$.

Concept Coulomb's law - force and resulting acceleration

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42 questions from this chapter have appeared across 23 exam years.