Question

Four charges each equal to $Q$ are placed at the four corners
of a square and a charge $q$ is placed at its center of the square. If the system
is in equilibrium then the value of $q$ is

• A. $-\frac{Q}{4}(1+2\sqrt{2})$
• B. $\frac{Q}{4}(1+2\sqrt{2})$
• C. $-\frac{Q}{2}(1+2\sqrt{2})$
• D. $\frac{Q}{2}(1+2\sqrt{2})$

Answer 1

The correct option is A $-\frac{Q}{4}(1+2\sqrt{2})$

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

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

From the figure,

$BD=\sqrt{2}a$

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

Now, the force is

$F_{BA}=F_{BC}=\frac{Q^2}{4\pi {\epsilon }_0{a}^2}$

$F_{BD}=\frac{Q^2}{4\pi {\epsilon }_0{\left(\sqrt{2}a\right)}^2}$

$F_{BO}=\frac{qQ}{4\pi {\epsilon }_0{\left(\frac{a}{\sqrt{2}}\right)}^2}$

$F_{BO}=\frac{-2qQ}{4\pi {\epsilon }_0{a}^2}$

The net force acting on –Q at point will be zero if

$F_{BA}\cos {45}^0+F_{BC}\cos {45}^0+F_{BD}+F_{BO}=0$

$\frac{Q^2}{4\pi {\epsilon }_0{a}^2}\times \frac{1}{\sqrt{2}}+\frac{Q^2}{4\pi {\epsilon }_0{a}^2}\times \frac{1}{\sqrt{2}}+\frac{Q^2}{4\pi {\epsilon }_0{\left(2a\right)}^2}-\frac{2qQ}{4\pi {\epsilon }_0{a}^2}=0$

After solving

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

Hence, the value of q is $\frac{Q}{4}(1+2\sqrt{2})$