Question

At what distance of separation must two $1\mu C$ charges be positioned in order for the repulsive force between them to be equivalent to the weight (on Earth) of a $1kg$ mass?

Answer 1

Hint:This question utilizes the concept of force experienced by two charged bodies and force by gravity. We first find the force experienced by $1kg$ mass on earth and then we substitute that force in coulombs formula for force exerted by charged bodies on each other.

Formulae used :
\[{F_c} = k\dfrac{{{Q_1}{Q_2}}}{{{r^2}}}\] where ${F_c}$ is the coulomb’s force exerted by both bodies on each other, $k$ is the coulomb’s constant, ${Q_1}$ is the charge on the first body, ${Q_2}$ is the charge on the second body and $r$ is the distance between the two charges.
$F = mg$ where $F$ is the force, $m$ is the mass of the body and $g$ is the acceleration due to gravity
Acceleration due to gravity $g = 9.8m{s^{ - 2}}$
Coulomb’s constant $k = 9 \times {10^9}$

Complete step by step answer:
The weight of a body $A$ of mass $1kg$ on earth will be
$
   \Rightarrow F = m \times g \\
   \Rightarrow F = 1kg \times 9.8m{s^{ - 2}} \\
 $
$ \Rightarrow F = 9.8N$ --------------------------(i)
Now, we are told that the repulsive force should be equal to this force $F$ . Thus, we have
$ \Rightarrow {F_c} = F$
From eq(i), we get
${F_c} = 9.8N$
Now, we also know that repulsive force between two charges will be
 \[ \Rightarrow {F_c} = k\dfrac{{{Q_1}{Q_2}}}{{{r^2}}}\]
Substituting the respective values, we get
\[ \Rightarrow 9.8 = \dfrac{{\left( {9{\kern 1pt} \times {{10}^9}} \right){{\left( {1 \times {{10}^{ - 6}}} \right)}^2}}}{{{r^2}}}\]
\[ \Rightarrow {r^2} = \dfrac{{\left( {9{\kern 1pt} \times {{10}^9}} \right){{\left( {1 \times {{10}^{ - 6}}} \right)}^2}}}{{9.8}}\]
Using square roots on both sides, we get
\[
   \Rightarrow r = \sqrt {\dfrac{{\left( {9{\kern 1pt} \times {{10}^9}} \right){{\left( {1 \times {{10}^{ - 6}}} \right)}^2}}}{{9.8}}} \\
   \Rightarrow r = 0.0303m \\
 \]
Therefore, required distance between the two charges will be $0.0303m$

Note: Students usually get confused when doing square roots. So, make sure that first you take out all the terms from the square root and then solve them. Also, we could have used acceleration due to gravity as $10m{s^{ - 2}}$ and we would’ve got the answer as $0.03m$ which is acceptable.