Question

Two charges each of +Q units are placed along a line. A third charge -q is placed between them. At what position and for what value of q, will the system be in equilibrium?

Answer 1

Hint: Draw the diagram showing charges +Q units placed along a line and the third charge placed between them. Assume the total distance between +Q units charges be r.
Formula used - \[F = \dfrac{{{Q_1}{Q_2}}}{{4\pi { \in _ \circ }{a^2}}}\]

Complete step by step answer:
Given that- Two charges each of +Q units are placed along a line, the third charge -q is placed between them.
Refer to the figure below-

Let the charge q be placed at O, so that its distance from charge at A be x and its distance from charge at B be (r – x).
Now, the electrostatic force on a charge due to other charge is given by the formula, \[F = \dfrac{{{Q_1}{Q_2}}}{{4\pi { \in _ \circ }{a^2}}}\]
Here Q1 and Q2 are the charges, F is the electrostatic force and a is the distance between the two charges.
So, electrostatic force on q due to +Q charge at A is \[\dfrac{{qQ}}{{4\pi { \in _ \circ }{x^2}}}\]
And, electrostatic force on q due to +Q charge at B is \[\dfrac{{qQ}}{{4\pi { \in _ \circ }{{(r - x)}^2}}}\]
Now, the system is in equilibrium, if electrostatic force on q due to charge at A = electrostatic force on q due to charge at B, i.e.,
\[\dfrac{{qQ}}{{4\pi { \in _ \circ }{x^2}}} = \dfrac{{qQ}}{{4\pi { \in _ \circ }{{(r - x)}^2}}}\]
Solving further we get-
$
  {(r - x)^2} = {x^2} \\
   \Rightarrow {r^2} + {x^2} - 2rx = {x^2} \\
   \Rightarrow {r^2} = 2rx \\
   \Rightarrow x = \dfrac{r}{2} \\
 $
Also, the three charges will be in equilibrium if net force on each charge is zero i.e.
\[
  \dfrac{{qQ}}{{4\pi { \in _ \circ }{{\left( {\dfrac{r}{2}} \right)}^2}}} = \dfrac{{QQ}}{{4\pi { \in _ \circ }{{(r)}^2}}} \\
   \Rightarrow \dfrac{q}{{4\pi { \in _ \circ }{{\left( {\dfrac{r}{2}} \right)}^2}}} = \dfrac{Q}{{4\pi { \in _ \circ }{{\left( r \right)}^2}}} \\
   \Rightarrow q = \dfrac{Q}{4} \\
 \]
Therefore, the third charge has magnitude equal to Q/4 and is placed between the two charges at centre.

Note – According to the Coulomb’s law, the electrostatic force on a charge due to other charge is given by the formula, \[F = \dfrac{{{Q_1}{Q_2}}}{{4\pi { \in _ \circ }{a^2}}}\] . Then, for charge -q to be in equilibrium, the force acting on -q due to +Q at A and +Q at B should be equal and opposite and to find the magnitude of the charge we have consider the case where charge +Q at A or at B are in equilibrium.