Question

Find the charge Q that should be placed at the centre of the square of side 7.10E+0 cm, at the corners of which four identical charges +q = 11.5 C are placed so that the whole system is in equilibrium

Find the charge O that should be placed at the centre of the square of side 7.10E+0 cm, at the corners of which four identical charges +9 = 11.5 C are placed so that the whole system is in equilibrium. 09------ Qa qÓ------

Answer 1

Let,

• $a$ : be the sides of the square.

Given:-

• $+q = 11.5 \: C$
• $\small a = 7.1 \: cm = 0.071 \: m$

Now, lets label the corners of the square and center of the square as O, as shown in the below figure,

nola

In order to have equilibrium in the system, charge Q must be negative.

Let,

• $\small AB = BC =CD=DA=a = 0.071 \: m$

From the figure,

$\small DB = \sqrt{\left ( a \right )^{2} + \left ( a \right )^{2} }$

$\small DB = \sqrt{2} \: a$

Therefore,

$\small OB = \frac{1}{2} DB = \frac{1}{2} \sqrt{2} \left (a \right )$

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

By Coulomb's law, force between the two charges is given as,

$F =\left ( \frac{1}{4 \pi \epsilon _{0}} \right ) \frac{q^{2}}{a^{2}}$

Charge at point B experience forces F_BA, F_BC and F_BD.

Forces F_BA and F_BC will be equal, therefore,

$F_{BA} =F_{BC}= \left ( \frac{1}{4 \pi \epsilon _{0}} \right ) \frac{q^{2}}{a^{2}}$

F_BD is given as,

$F_{BD} = \left ( \frac{1}{4 \pi \epsilon _{0}} \right ) \frac{q^{2}}{\left ( DB \right )^{2}}$

$F_{BD} = \left ( \frac{1}{4 \pi \epsilon _{0}} \right ) \frac{q^{2}}{\left ( \sqrt{2}a \right )^{2}}$

Charge at point B also experience force F_BO, and is given as,

$F_{BO} = \left ( \frac{1}{4 \pi \epsilon _{0}} \right ) \frac{q \left ( Q \right )}{\left ( OB \right )^{2}}$

$F_{BO} = \left ( \frac{1}{4 \pi \epsilon _{0}} \right ) \frac{q \left ( Q \right )}{\left ( \frac{a}{\sqrt{2}} \right )^{2}}$

$F_{BO} = \left ( \frac{1}{4 \pi \epsilon _{0}} \right ) \frac{2 q Q }{\left ( a \right )^{2}}$

We have just taken the magnitude of the force.

To maintain the equilibrium in a system, net force on the charge at point B will be zero if,

$F_{BA} \, \cos45 + F_{BC} \, \cos45 + F_{BD}+ F_{BO} = 0$

Since, $\inline F_{BA} = F_{BC}$ ,

$\inline 2 F_{BA} \, \cos45 + F_{BD}+ F_{BO} = 0$

$\frac{1}{4 \pi \epsilon _{0}} \left (\frac{2q^{2}}{ a^{2}} \right ) \, \cos45+\frac{1}{4 \pi \epsilon _{0}} \left (\frac{q^{2}}{\left ( \sqrt{2}a \right )^{2}} \right ) +\frac{1}{4 \pi \epsilon _{0}} \left (\frac{2qQ}{ a^{2}} \right ) = 0$

$\frac{1}{4 \pi \epsilon _{0}} \left (\frac{2q^{2}}{ a^{2}} \right ) \frac{1}{\sqrt{2}}+\frac{1}{4 \pi \epsilon _{0}} \left (\frac{q^{2}}{2 a^{2}} \right ) + \frac{1}{4 \pi \epsilon _{0}} \left (\frac{2qQ}{ a^{2}} \right ) = 0$

$\sqrt{2} \left (\frac{q^{2}}{ a^{2}} \right )+ \frac{1}{2} \left (\frac{q^{2}}{ a^{2}} \right ) + 2 \left (\frac{qQ}{ a^{2}} \right ) = 0$

$\sqrt{2} \left (q^{2} \right )+ \frac{1}{2} \left (q^{2} \right ) + 2 \left (qQ \right ) = 0$

$\sqrt{2} \left (q \right )+ \frac{1}{2} \left (q \right ) + 2 \left (Q \right ) = 0$

$q \left (\sqrt{2}+ \frac{1}{2} \right ) + 2 \left (Q \right ) = 0$

$2 \left (Q \right ) = -q \left (\sqrt{2}+ \frac{1}{2} \right )$

$2 \left (Q \right ) =- q \frac{\left (2 \sqrt{2}+ 1 \right ) }{2}$

$\mathbf{Q = - \frac{q}{4} \left (2 \sqrt{2}+ 1 \right )}$

This will be the value of charge Q at the center of the square such that the system will be in equilibrium.