Question

A charge q is placed at every corner of a square having side a. how much charge must be placed at its center to bring the system in equilibrium?

Answer 1

Step 1: Given data:

A figure to represent the given information:

If each charge shows no force then the system is in equilibrium.

In the figure, it is shown that the charge ‘q’ at the center would be in equilibrium for any value of q.

Now, let $AB=BC=CD=DA=a$

We get from the figure that,

$DB=\sqrt{2}$and $OB=\frac{a}{\sqrt{2}}$

Step 2: Calculation:

So, $F_{BA}=F_{BC}=\frac{Q^2}{4\mathrm{\pi }\in \circ {\mathrm{a}}^2}$

$$\begin{gathered} F_{BD}=\frac{Q^2}{4\mathrm{\pi }\in \circ {\left(\sqrt{2}\mathrm{a}\right)}^2}=\frac{Q^2}{4\mathrm{\pi }\in \circ {\left(2\mathrm{a}\right)}^2} \\ {F}_{BD}=\frac{qQ}{4\mathrm{\pi }\in \circ {\left({\displaystyle \frac{a}{\sqrt{2}}}\right)}^2}=\frac{2qQ}{4\mathrm{\pi }\in \circ a^2} \end{gathered}$$

And, the net force on Q at point B will be zero if

$F_{BA}\cos 45+F_{Bc}cos45+F_{BD}+F_{BO}=0$

Then, $\frac{Q^2}{4\mathrm{\pi }\in \circ {\mathrm{a}}^2}\times \frac{1}{\sqrt{2}}+\frac{Q^2}{4\mathrm{\pi }\in \circ {\mathrm{a}}^2}\times \frac{1}{\sqrt{2}}+\frac{Q^2}{4\mathrm{\pi }\in \circ {\left(2\mathrm{a}\right)}^2}-\frac{2qQ}{4\mathrm{\pi }\in \circ {\mathrm{a}}^2}=0$

On solving the equation,

$q=\frac{Q}{4}(1+2\sqrt{2})$

Hence, the charge to be placed at its center to bring the system to equilibrium is $q=\frac{Q}{4}(1+2\sqrt{2})$.