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# The value on net force on charge ${Q_0}$

Last updated date: 07th Sep 2024
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Hint: In this question we will use the relation of electrostatic force between the charges and the distance between them. Also, by applying the formula and calculating step by step, we get the required result. At last we will discuss the basics of electrostatic force and also mechanical force.
Formula used:
$F = \dfrac{{kQ{Q_0}}}{{{a^2}}}$

Here we can know that the force due to charge Q at D and B along DB is given as:
$F' = {F_D} + {F_B}$
As we know, force is given by:
$F' = \dfrac{{kQ{Q_0}}}{{{{\left( {a\sqrt 2 } \right)}^2}}} + \dfrac{{k( + Q){Q_0}}}{{{{\left( {a\sqrt 2 } \right)}^2}}}$
Here, Q and ${Q_0}$ are charges, a is distance between the two charges, k is known as Coulomb’s law constant having value $9 \times {10^9}N{m^2}/{C^2}$
Further, solving we get:
\eqalign{& \Rightarrow F' = 2 \times 2\dfrac{{kQ{Q_0}}}{{{a^2}}} \cr & \Rightarrow F' = 4\dfrac{{kQ{Q_0}}}{{{a^2}}} \cr}
Similarly the force along CA, due to charge Q and ${Q_0}$ at A and C will be given as:
$F'' = {F_A} + {F_C}$
$\Rightarrow F'' = 4\dfrac{{kQ{Q_0}}}{{{a^2}}}$
Now, resultant force will be along OM, which is given as:
${F_R} = \sqrt {{{(F')}^2} + {{(F'')}^2} + 2(F'F'')cos{{90}^ \circ }}$
By substituting the value of forces in above equation, we get:
\eqalign{& \Rightarrow {F_R} = \sqrt {{{\left( {4\dfrac{{kQ{Q_0}}}{{{a^2}}}} \right)}^2} + {{\left( {4\dfrac{{kQ{Q_0}}}{{{a^2}}}} \right)}^2} + 0} \cr & \Rightarrow {F_R} = \sqrt 2 \times 4\dfrac{{kQ{Q_0}}}{{{a^2}}} \cr & \therefore {F_R} = 4\sqrt 2 \dfrac{{kQ{Q_0}}}{{{a^2}}} \cr}
Therefore, we get the required result in above equation, which gives expression of net force on charge ${Q_0}$