Answer
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Hint: The above problem can be resolved using the mathematical formula for the electrostatic force, acing between the two charged species and the formula for the weight of the particle. In the given problem, it is said that while the electrostatic force is acting within the charged particle, its magnitude is equal to the weight of the particle. Then by taking the numerical comparison for the electrostatic force and the weight, the desired value for the separation between the particles is calculated.
Complete step by step answer:
Given:
The mass of each charged particle is, \[m = 1.7 \times {10^{ - 27}}\;{\rm{kg}}\].
The magnitude of charge is, \[q = 1.6 \times {10^{ - 19}}\;{\rm{C}}\].
We know the magnitude of electrostatic force is,
\[{F_1} = \dfrac{{k{q^2}}}{{{d^2}}}\]…… (1)
Here, k is the coulomb’s constant and its value is \[9 \times {10^9}\;{\rm{N/C}}\] and d is the separating distance.
And the expression for the weight is given by,
\[W = mg\]…….. (2)
Here, g is the gravitational acceleration and its value is \[9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\].
As the experience is equivalent to the weight. Hence comparing the equations 1 and 2 as,
\[\begin{array}{l}
{F_1} = W\\
\dfrac{{k{q^2}}}{{{d^2}}} = mg\\
d = \sqrt {\dfrac{{k{q^2}}}{{mg}}}
\end{array}\]
Now, substituting the values in the above equation as,
\[\begin{array}{l}
d = \sqrt {\dfrac{{k{q^2}}}{{mg}}} \\
d = \sqrt {\dfrac{{9 \times {{10}^9}\;{\rm{N/C}} \times {{\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C}}} \right)}^2}}}{{1.7 \times {{10}^{ - 27}}\;{\rm{kg}} \times 9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}}}} \\
d = 0.11\;{\rm{m}}
\end{array}\]
Therefore, the value of d is \[0.11\;{\rm{m}}\].
Note: To resolve the given problem, one must be clear about the concepts involved in the balancing of forces. Here in the given problem, the forces that are supposed to be in balanced condition are namely the electrostatic force and the magnitude of force due to gravity. And the magnitude of force due to gravity is known as weight. Moreover, it is important to remember the formulas of these forces to proceed with the final solution.
Complete step by step answer:
Given:
The mass of each charged particle is, \[m = 1.7 \times {10^{ - 27}}\;{\rm{kg}}\].
The magnitude of charge is, \[q = 1.6 \times {10^{ - 19}}\;{\rm{C}}\].
We know the magnitude of electrostatic force is,
\[{F_1} = \dfrac{{k{q^2}}}{{{d^2}}}\]…… (1)
Here, k is the coulomb’s constant and its value is \[9 \times {10^9}\;{\rm{N/C}}\] and d is the separating distance.
And the expression for the weight is given by,
\[W = mg\]…….. (2)
Here, g is the gravitational acceleration and its value is \[9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\].
As the experience is equivalent to the weight. Hence comparing the equations 1 and 2 as,
\[\begin{array}{l}
{F_1} = W\\
\dfrac{{k{q^2}}}{{{d^2}}} = mg\\
d = \sqrt {\dfrac{{k{q^2}}}{{mg}}}
\end{array}\]
Now, substituting the values in the above equation as,
\[\begin{array}{l}
d = \sqrt {\dfrac{{k{q^2}}}{{mg}}} \\
d = \sqrt {\dfrac{{9 \times {{10}^9}\;{\rm{N/C}} \times {{\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C}}} \right)}^2}}}{{1.7 \times {{10}^{ - 27}}\;{\rm{kg}} \times 9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}}}} \\
d = 0.11\;{\rm{m}}
\end{array}\]
Therefore, the value of d is \[0.11\;{\rm{m}}\].
Note: To resolve the given problem, one must be clear about the concepts involved in the balancing of forces. Here in the given problem, the forces that are supposed to be in balanced condition are namely the electrostatic force and the magnitude of force due to gravity. And the magnitude of force due to gravity is known as weight. Moreover, it is important to remember the formulas of these forces to proceed with the final solution.
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