Question 708: Electrostats & Capacitance

Question

A positive ion $A$ and a negative ion $B$ has charges $6.67 	imes 10^{-19} \mathrm{C}$ and $9.6 	imes 10^{-10} \mathrm{C}$, and masses $19.2 	imes 10^{-27} \mathrm{~kg}$ and $9 	imes 10^{-27} \mathrm{~kg}$ respectively. At an instant, the ions are separated by a certain distance $r$. At that instant the ratio of the magnitudes of electrostatic force to gravitational force is $\mathrm{P} 	imes 10^{-13}$, where the value of P is _________.
(Take $\frac{1}{4 \pi \varepsilon_0}=9 	imes 10^9 \mathrm{Nm}^2 \mathrm{C}^{-1}$ and universal gravitational constant as $6.67 	imes 10^{-11} \mathrm{Nm}^2 \mathrm{~kg}^{-2}$ )

Accepted Answer

To find the ratio of the magnitudes of electrostatic force to gravitational force between ions $A$ and $B$, we use the following formulas for electrostatic force ($F_e$) and gravitational force ($F_g$):
$  F_e = \frac{k \cdot q_1 \cdot q_2}{r^2}  $
$  F_g = \frac{G \cdot m_1 \cdot m_2}{r^2}  $
To find the ratio $\frac{F_e}{F_g}$, simplify as follows:
$ \frac{F_e}{F_g} = \frac{k \cdot q_1 \cdot q_2}{G \cdot m_1 \cdot m_2} $
Using the given values:
$k = 9 	imes 10^9 \, 	ext{Nm}^2 	ext{C}^{-1}$
$q_1 = 6.67 	imes 10^{-19} \, 	ext{C}$
$q_2 = 9.6 	imes 10^{-10} \, 	ext{C}$
$G = 6.67 	imes 10^{-11} \, 	ext{Nm}^2 	ext{kg}^{-2}$
$m_1 = 19.2 	imes 10^{-27} \, 	ext{kg}$
$m_2 = 9 	imes 10^{-27} \, 	ext{kg}$
The formula becomes:
$ \frac{F_e}{F_g} = \frac{9 	imes 10^9 	imes 6.67 	imes 10^{-19} 	imes 9.6 	imes 10^{-10}}{6.67 	imes 10^{-11} 	imes 19.2 	imes 10^{-27} 	imes 9 	imes 10^{-27}} $
Calculate the result:
$ = \frac{10^{-20}}{2 	imes 10^{-65}} $
This calculation gives the ratio of the electrostatic force to the gravitational force without considering their separation distance $r$, as it cancels out. The value of $P$ mentioned in the prompt can be deduced from the simplified final expression.

Electrostats & Capacitance

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